Codeforces Round #290 (Div. 2)

评测之前

[remind]总结：看到Div1的A题我就吓尿了、然后换小号做了Div2、最后1秒交了Div2的D题，然后Present Pass，我也是醉了，果然我太弱了、、然后没有了、、、[/remind]

好吧，那么让我们来分析题目>_<

[task]

Fox Ciel starts to learn programming. The first task is drawing a fox! However, that turns out to be too hard for a beginner, so she decides to draw a snake instead.
A snake is a pattern on a n by m table. Denote c-th cell of r-th row as (r, c). The tail of the snake is located at (1, 1), then it’s body extends to (1, m), then goes down 2 rows to (3, m), then goes left to (3, 1) and so on.
Your task is to draw this snake for Fox Ciel: the empty cells should be represented as dot characters (‘.‘) and the snake cells should be filled with number signs (‘#‘).
Consider sample tests in order to understand the snake pattern.

Input

The only line contains two integers: n and m (3 ≤ n, m ≤ 50).
n is an odd number.

Output

Output n lines. Each line should contain a string consisting of m characters. Do not output spaces.

Sample test(s)

input

3 3

3 3

output

###
..#
###

###

..#

###

input

3 4

3 4

output

####
...#
####

####

...#

####

input

5 3

5 3

output

###
..#
###
#..
###

###

..#

###

#..

###

input

9 9

9 9

output

#########
........#
#########
#........
#########
........#
#########
#........
#########

#########

........#

#########

#........

#########

........#

#########

#........

#########

[/task]
题目大意大概就是、、、卧槽、好吧、模拟、、看数据就能知道辣！

/*
  Author:WNJXYK
  WebSite:http://wnjxyk.cn
  Weibo:http://weibo.com/WNJXYK
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m;
int tmp=0;
int main(){
    scanf("%d%d",&n,&m);
    for (int i=1;i<=n;i++){
        if (i&1){
            for (int j=1;j<=m;j++)printf("#");
        }else{
            ++tmp;
            if (tmp&1){
                for (int j=1;j<m;j++) printf(".");
                printf("#");
            }else{
                printf("#");
                for (int j=1;j<m;j++) printf(".");
            }
        }
        printf("n");
    }
    return 0;
}

Author:WNJXYK

WebSite:http://wnjxyk.cn

Weibo:http://weibo.com/WNJXYK

#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

int n,m;

int tmp=0;

int main(){

scanf("%d%d",&n,&m);

for (int i=1;i<=n;i++){

if (i&1){

for (int j=1;j<=m;j++)printf("#");

}else{

++tmp;

if (tmp&1){

for (int j=1;j<m;j++) printf(".");

printf("#");

}else{

printf("#");

for (int j=1;j<m;j++) printf(".");

}

printf("n");

}

return 0;

}

[remind]

Fox Ciel is playing a mobile puzzle game called “Two Dots”. The basic levels are played on a board of size n × m cells, like this:
Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.
The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots d₁, d₂, …, d_k a cycle if and only if it meets the following condition:

These k dots are different: if i ≠ j then d_i is different from d_j.
k is at least 4.
All dots belong to the same color.
For all 1 ≤ i ≤ k - 1: d_i and d_i + 1 are adjacent. Also, d_k and d₁ should also be adjacent. Cells x and y are called adjacent if they share an edge.

Determine if there exists a cycle on the field.

Input

The first line contains two integers n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.
Then n lines follow, each line contains a string consisting of m characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.

Output

Output “Yes” if there exists a cycle, and “No” otherwise.

Sample test(s)

input

3 4
AAAA
ABCA
AAAA

3 4

AAAA

ABCA

AAAA

output

Yes

Yes

input

3 4
AAAA
ABCA
AADA

3 4

AAAA

ABCA

AADA

output

No

input

4 4
YYYR
BYBY
BBBY
BBBY

4 4

YYYR

BYBY

BBBY

output

Yes

Yes

input

7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB

7 6

AAAAAB

ABBBAB

ABAAAB

ABABBB

ABAAAB

ABBBAB

AAAAAB

output

Yes

Yes

input

2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ

2 13

ABCDEFGHIJKLM

NOPQRSTUVWXYZ

output

No

Note

In first sample test all ‘A‘ form a cycle.
In second sample there is no such cycle.
The third sample is displayed on the picture above (‘Y‘ = Yellow, ‘B‘ = Blue, ‘R‘ = Red).

[/remind]
题目大意就是我们要查找是否有长度大于4的环

/*
  Author:WNJXYK
  WebSite:http://wnjxyk.cn
  Weibo:http://weibo.com/WNJXYK
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int Maxn=55;
int n,m;
char map[Maxn][Maxn];
string intmp;
bool vis[Maxn][Maxn];
int ind[Maxn][Maxn];
int Ans;
int dx[]={0,1,0,-1};
int dy[]={1,0,-1,0};
inline int remax(int a,int b){
    if (a>b) return a;
    return b;
}
inline int zabs(int x){
    if (x<0) return -x;
    return x;
}
void dfs(int x,int y,int st){
    vis[x][y]=true;
    ind[x][y]=st;
    for (int k=0;k<4;k++){
        int nx=x+dx[k],ny=y+dy[k];
        if (!(1<=nx && nx<=n && 1<=ny && ny<=m))continue;
        if (map[nx][ny]==map[x][y]){
            if (vis[nx][ny])
                Ans=remax(Ans,zabs(ind[x][y]-ind[nx][ny]));
            else
                dfs(nx,ny,st+1);
        }
    }
}
int main(){
    scanf("%d%d",&n,&m);
    for (int i=1;i<=n;i++){
        cin>>intmp;
        for (int j=1;j<=m;j++){
            map[i][j]=intmp[j-1];
        }
    }
    for (int i=1;i<=n;i++){
        for (int j=1;j<=m;j++){
            if(!vis[i][j])dfs(i,j,0);
            if (Ans>=3) {
                printf("Yesn");
                return 0;
            }
        }
    }
    printf("Non");
    return 0;
}

Author:WNJXYK

WebSite:http://wnjxyk.cn

Weibo:http://weibo.com/WNJXYK

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

const int Maxn=55;

int n,m;

char map[Maxn][Maxn];

string intmp;

bool vis[Maxn][Maxn];

int ind[Maxn][Maxn];

int Ans;

int dx[]={0,1,0,-1};

int dy[]={1,0,-1,0};

inline int remax(int a,int b){

if (a>b) return a;

return b;

}

inline int zabs(int x){

if (x<0) return -x;

return x;

}

void dfs(int x,int y,int st){

vis[x][y]=true;

ind[x][y]=st;

for (int k=0;k<4;k++){

int nx=x+dx[k],ny=y+dy[k];

if (!(1<=nx && nx<=n && 1<=ny && ny<=m))continue;

if (map[nx][ny]==map[x][y]){

if (vis[nx][ny])

Ans=remax(Ans,zabs(ind[x][y]-ind[nx][ny]));

else

dfs(nx,ny,st+1);

}

int main(){

scanf("%d%d",&n,&m);

for (int i=1;i<=n;i++){

cin>>intmp;

for (int j=1;j<=m;j++){

map[i][j]=intmp[j-1];

}

for (int i=1;i<=n;i++){

for (int j=1;j<=m;j++){

if(!vis[i][j])dfs(i,j,0);

if (Ans>=3) {

printf("Yesn");

return 0;

}

printf("Non");

return 0;

}

[remind]

Fox Ciel is going to publish a paper on FOCS (Foxes Operated Computer Systems, pronounce: “Fox”). She heard a rumor: the authors list on the paper is always sorted in the lexicographical order.
After checking some examples, she found out that sometimes it wasn’t true. On some papers authors’ names weren’t sorted inlexicographical order in normal sense. But it was always true that after some modification of the order of letters in alphabet, the order of authors becomes lexicographical!
She wants to know, if there exists an order of letters in Latin alphabet such that the names on the paper she is submitting are following in the lexicographical order. If so, you should find out any such order.
Lexicographical order is defined in following way. When we compare s and t, first we find the leftmost position with differing characters:s_i ≠ t_i. If there is no such position (i. e. s is a prefix of t or vice versa) the shortest string is less. Otherwise, we compare characters s_iand t_i according to their order in alphabet.

Input

The first line contains an integer n (1 ≤ n ≤ 100): number of names.
Each of the following n lines contain one string name_i (1 ≤ |name_i| ≤ 100), the i-th name. Each name contains only lowercase Latin letters. All names are different.

Output

If there exists such order of letters that the given names are sorted lexicographically, output any such order as a permutation of characters ‘a’–’z’ (i. e. first output the first letter of the modified alphabet, then the second, and so on).
Otherwise output a single word “Impossible” (without quotes).

Sample test(s)

input

3
rivest
shamir
adleman

rivest

shamir

adleman

output

bcdefghijklmnopqrsatuvwxyz

1	bcdefghijklmnopqrsatuvwxyz

input

10
tourist
petr
wjmzbmr
yeputons
vepifanov
scottwu
oooooooooooooooo
subscriber
rowdark
tankengineer

tourist

petr

wjmzbmr

yeputons

vepifanov

scottwu

oooooooooooooooo

subscriber

rowdark

tankengineer

output

Impossible

1	Impossible

input

10
petr
egor
endagorion
feferivan
ilovetanyaromanova
kostka
dmitriyh
maratsnowbear
bredorjaguarturnik
cgyforever

petr

egor

endagorion

feferivan

ilovetanyaromanova

kostka

dmitriyh

maratsnowbear

bredorjaguarturnik

cgyforever

output

aghjlnopefikdmbcqrstuvwxyz

1	aghjlnopefikdmbcqrstuvwxyz

input

7
car
care
careful
carefully
becarefuldontforgetsomething
otherwiseyouwillbehacked
goodluck

car

care

careful

carefully

becarefuldontforgetsomething

otherwiseyouwillbehacked

goodluck

output

acbdefhijklmnogpqrstuvwxyz

1	acbdefhijklmnogpqrstuvwxyz

[/remind]
题目大意就是我们有一系列假装已经排好序的字符串！我们要更改abcdedfhigklmnopqrstuvwxyz的大小顺序，来使这些字符串成字典序。
我当时感觉就是小的到大的连边，然后做一遍SPFA，然后按照点的权值输出就好！不知道会不会fst、、、

/*
  Author:WNJXYK
  WebSite:http://wnjxyk.cn
  Weibo:http://weibo.com/WNJXYK
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<queue>
using namespace std;
const int Maxn=105;
int n;
string name[Maxn];
bool map[30][30];
struct Edge{
    int v;
    int nxt;
    Edge(){
    }
    Edge(int a,int b){
        v=a;nxt=b;
    }
};
int head[30];
int nume;
Edge e[Maxn*200];
inline void addEdge(int u,int v){
    e[++nume]=Edge(v,head[u]);
    head[u]=nume;
}
inline bool compare(int x,int y){
    int len=name[x].length()<name[y].length()?name[x].length():name[y].length();
    //cout<<name[x]<<" "<<name[y]<<" => "<<endl;
    for (int i=0;i<len;i++){
        if (name[x][i]!=name[y][i]){
            int a=name[x][i]-'a'+1,b=name[y][i]-'a'+1;
            if (map[a][b]==false){
                map[a][b]=true;
                addEdge(a,b);
                //printf("%d %c - %cn",i,name[x][i],name[y][i]);
                return true;
            }else{
                return true;
            }
        }else{
            //cout<<name[x][i]<<"="<<name[y][i]<<endl;
        }
    }
    if (name[x].length()>name[y].length()) return false;
    return true;
}
queue<int> que;
struct Dist{
    int w;
    char x;
};
Dist dist[30];
bool inque[30];
int inques[30];
inline bool spfa(){
    dist[1].w=30;
    for (int i=2;i<=26;i++)  dist[i].w=dist[i-1].w+30;
    for (int i=1;i<=26;i++) dist[i].x=(char)('a'+i-1);
    while(!que.empty()) que.pop();
    for (int i=1;i<=26;i++){
        que.push(i);
        inque[i]=true;
        inques[i]=0;
    }
    while(!que.empty()){
        int now=que.front();que.pop();
        for (int i=head[now];i;i=e[i].nxt){
            int v=e[i].v;
            if (dist[v].w<dist[now].w+1){
                dist[v].w=dist[now].w+1;
                if (inque[v]==false){
                    inques[v]++;
                    if (inques[v]>n) return false;
                    inque[v]=true;
                    que.push(v);
                }
            }
        }
        inque[now]=false;
    }
    return true;
}
bool cmp(Dist a,Dist b){
    if (a.w<b.w) return true;
    return false;
}
int main(){
    scanf("%d",&n);
    for (int i=1;i<=n;i++) cin>>name[i];
    for (int i=1;i<n;i++){
        for (int j=i+1;j<=n;j++){
            if (!compare(i,j)){
                printf("Impossiblen");
                return 0;
            }
        }
    }
    if (spfa()){
        sort(dist+1,dist+27,cmp);
        for (int i=1;i<=26;i++) cout<<dist[i].x;
        cout<<endl;
    }else{
        printf("Impossiblen");
    }
    return 0;
}

100

101

102

103

104

105

106

107

108

109

110

111

112

113

Author:WNJXYK

WebSite:http://wnjxyk.cn

Weibo:http://weibo.com/WNJXYK

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<string>

#include<queue>

using namespace std;

const int Maxn=105;

int n;

string name[Maxn];

bool map[30][30];

struct Edge{

int v;

int nxt;

Edge(){

}

Edge(int a,int b){

v=a;nxt=b;

}

};

int head[30];

int nume;

Edge e[Maxn*200];

inline void addEdge(int u,int v){

e[++nume]=Edge(v,head[u]);

head[u]=nume;

}

inline bool compare(int x,int y){

int len=name[x].length()<name[y].length()?name[x].length():name[y].length();

//cout<<name[x]<<" "<<name[y]<<" => "<<endl;

for (int i=0;i<len;i++){

if (name[x][i]!=name[y][i]){

int a=name[x][i]-'a'+1,b=name[y][i]-'a'+1;

if (map[a][b]==false){

map[a][b]=true;

addEdge(a,b);

//printf("%d %c - %cn",i,name[x][i],name[y][i]);

return true;

}else{

return true;

}

}else{

//cout<<name[x][i]<<"="<<name[y][i]<<endl;

}

if (name[x].length()>name[y].length()) return false;

return true;

}

queue<int> que;

struct Dist{

int w;

char x;

};

Dist dist[30];

bool inque[30];

int inques[30];

inline bool spfa(){

dist[1].w=30;

for (int i=2;i<=26;i++) dist[i].w=dist[i-1].w+30;

for (int i=1;i<=26;i++) dist[i].x=(char)('a'+i-1);

while(!que.empty()) que.pop();

for (int i=1;i<=26;i++){

que.push(i);

inque[i]=true;

inques[i]=0;

}

while(!que.empty()){

int now=que.front();que.pop();

for (int i=head[now];i;i=e[i].nxt){

int v=e[i].v;

if (dist[v].w<dist[now].w+1){

dist[v].w=dist[now].w+1;

if (inque[v]==false){

inques[v]++;

if (inques[v]>n) return false;

inque[v]=true;

que.push(v);

}

inque[now]=false;

}

return true;

}

bool cmp(Dist a,Dist b){

if (a.w<b.w) return true;

return false;

}

int main(){

scanf("%d",&n);

for (int i=1;i<=n;i++) cin>>name[i];

for (int i=1;i<n;i++){

for (int j=i+1;j<=n;j++){

if (!compare(i,j)){

printf("Impossiblen");

return 0;

}

if (spfa()){

sort(dist+1,dist+27,cmp);

for (int i=1;i<=26;i++) cout<<dist[i].x;

cout<<endl;

}else{

printf("Impossiblen");

}

return 0;

}

[remind]

Fox Ciel is playing a game. In this game there is an infinite long tape with cells indexed by integers (positive, negative and zero). At the beginning she is standing at the cell 0.
There are also n cards, each card has 2 attributes: length l_i and cost c_i. If she pays c_i dollars then she can apply i-th card. After applying i-th card she becomes able to make jumps of length l_i, i. e. from cell x to cell (x - l_i) or cell (x + l_i).
She wants to be able to jump to any cell on the tape (possibly, visiting some intermediate cells). For achieving this goal, she wants to buy some cards, paying as little money as possible.
If this is possible, calculate the minimal cost.

Input

The first line contains an integer n (1 ≤ n ≤ 300), number of cards.
The second line contains n numbers l_i (1 ≤ l_i ≤ 10⁹), the jump lengths of cards.
The third line contains n numbers c_i (1 ≤ c_i ≤ 10⁵), the costs of cards.

Output

If it is impossible to buy some cards and become able to jump to any cell, output -1. Otherwise output the minimal cost of buying such set of cards.

Sample test(s)

input

3
100 99 9900
1 1 1

100 99 9900

1 1 1

output

2

input

5
10 20 30 40 50
1 1 1 1 1

10 20 30 40 50

1 1 1 1 1

output

-1

-1

input

7
15015 10010 6006 4290 2730 2310 1
1 1 1 1 1 1 10

15015 10010 6006 4290 2730 2310 1

1 1 1 1 1 1 10

output

6

input

8
4264 4921 6321 6984 2316 8432 6120 1026
4264 4921 6321 6984 2316 8432 6120 1026

4264 4921 6321 6984 2316 8432 6120 1026

output

7237

7237

Note

In first sample test, buying one card is not enough: for example, if you buy a card with length 100, you can’t jump to any cell whose index is not a multiple of 100. The best way is to buy first and second card, that will make you be able to jump to any cell.
In the second sample test, even if you buy all cards, you can’t jump to any cell whose index is not a multiple of 10, so you should output -1.

[/remind]
题目大意：有n个数字，每个数字有花费，我们要花最少的钱，是选择的数字的总GCD为1.
Orz一开始被忽悠过去做E、、【他们说E简单、、】然后最后十几分钟赶快写了一个纯暴力DP的D、【最后一秒钟提交Present Pass我也是醉、、、】总之很惊险，可能会T掉、【因为太暴力了】祝好运！

/*
  Author:WNJXYK
  WebSite:http://wnjxyk.cn
  Weibo:http://weibo.com/WNJXYK
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
int gcd(int x,int y){
    return y==0?x:gcd(y,x%y);
}
const int Maxn=305;
int cst[Maxn];
int len[Maxn];
int n;
map<int,int> dp[Maxn];
map<int,int>::iterator it;
int main(){
    scanf("%d",&n);
    for (int i=1;i<=n;i++) scanf("%d",&len[i]);
    for (int i=1;i<=n;i++) scanf("%d",&cst[i]);
    int tmpAllGcd=len[1];
    for (int i=2;i<=n;i++) tmpAllGcd=gcd(tmpAllGcd,len[i]);
    if (tmpAllGcd!=1){printf("-1n");return 0;}
    dp[1][len[1]]=cst[1];
    for (int i=2;i<=n;i++){
        dp[i][len[i]]=cst[i];
        for (it=dp[i-1].begin();it!=dp[i-1].end();it++){
            dp[i][it->first]=dp[i-1][it->first];
            int lenx=gcd(len[i],it->first);
            int val=it->second;
            if (dp[i][lenx]==0 || dp[i][lenx]>val+cst[i]) dp[i][lenx]=val+cst[i];
        }
    }
    int Ans=2147483647;
    for (int i=1;i<=n;i++) if (dp[i][1]!=0) Ans=Ans>dp[i][1]?dp[i][1]:Ans;
    printf("%dn",Ans);
    return 0;
}

Author:WNJXYK

WebSite:http://wnjxyk.cn

Weibo:http://weibo.com/WNJXYK

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<map>

using namespace std;

int gcd(int x,int y){

return y==0?x:gcd(y,x%y);

}

const int Maxn=305;

int cst[Maxn];

int len[Maxn];

int n;

map<int,int> dp[Maxn];

map<int,int>::iterator it;

int main(){

scanf("%d",&n);

for (int i=1;i<=n;i++) scanf("%d",&len[i]);

for (int i=1;i<=n;i++) scanf("%d",&cst[i]);

int tmpAllGcd=len[1];

for (int i=2;i<=n;i++) tmpAllGcd=gcd(tmpAllGcd,len[i]);

if (tmpAllGcd!=1){printf("-1n");return 0;}

dp[1][len[1]]=cst[1];

for (int i=2;i<=n;i++){

dp[i][len[i]]=cst[i];

for (it=dp[i-1].begin();it!=dp[i-1].end();it++){

dp[i][it->first]=dp[i-1][it->first];

int lenx=gcd(len[i],it->first);

int val=it->second;

if (dp[i][lenx]==0 || dp[i][lenx]>val+cst[i]) dp[i][lenx]=val+cst[i];

}

int Ans=2147483647;

for (int i=1;i<=n;i++) if (dp[i][1]!=0) Ans=Ans>dp[i][1]?dp[i][1]:Ans;

printf("%dn",Ans);

return 0;

}

[remind]E题没做(-｡-;)、、[/remind]

评测之后

A题Accept
B题Accept
C题Accept
D题FST、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、、世界再见、
好吧，昨天一躺下来就知道D哪里错了、、有一个更新的时候忘记判断了、、毕竟是15分钟码完然后最后1秒教的程序、、、、的确坑、、改了一下果然A掉了、QAQ、、SB错毁一生、

/*
  Author:WNJXYK
  WebSite:http://wnjxyk.cn
  Weibo:http://weibo.com/WNJXYK
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
int gcd(int x,int y){
    return y==0?x:gcd(y,x%y);
}
const int Maxn=305;
int cst[Maxn];
int len[Maxn];
int n;
map<int,int> dp[Maxn];
map<int,int>::iterator it;
int main(){
    scanf("%d",&n);
    for (int i=1;i<=n;i++) scanf("%d",&len[i]);
    for (int i=1;i<=n;i++) scanf("%d",&cst[i]);
    int tmpAllGcd=len[1];
    for (int i=2;i<=n;i++) tmpAllGcd=gcd(tmpAllGcd,len[i]);
    if (tmpAllGcd!=1){printf("-1n");return 0;}
    dp[1][len[1]]=cst[1];
    for (int i=2;i<=n;i++){
        dp[i][len[i]]=cst[i];
        for (it=dp[i-1].begin();it!=dp[i-1].end();it++){
            if (dp[i][it->first]==0 || dp[i][it->first]>dp[i-1][it->first])dp[i][it->first]=dp[i-1][it->first];
            //警察叔叔，就是上一句犯了SB错！
            int lenx=gcd(len[i],it->first);
            int val=it->second;
            if (dp[i][lenx]==0 || dp[i][lenx]>val+cst[i]) dp[i][lenx]=val+cst[i];
        }
    }
    int Ans=2147483647;
    for (int i=1;i<=n;i++) if (dp[i][1]!=0) Ans=Ans>dp[i][1]?dp[i][1]:Ans;
    printf("%dn",Ans);
    return 0;
}

Author:WNJXYK

WebSite:http://wnjxyk.cn

Weibo:http://weibo.com/WNJXYK

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<map>

using namespace std;

int gcd(int x,int y){

return y==0?x:gcd(y,x%y);

}

const int Maxn=305;

int cst[Maxn];

int len[Maxn];

int n;

map<int,int> dp[Maxn];

map<int,int>::iterator it;

int main(){

scanf("%d",&n);

for (int i=1;i<=n;i++) scanf("%d",&len[i]);

for (int i=1;i<=n;i++) scanf("%d",&cst[i]);

int tmpAllGcd=len[1];

for (int i=2;i<=n;i++) tmpAllGcd=gcd(tmpAllGcd,len[i]);

if (tmpAllGcd!=1){printf("-1n");return 0;}

dp[1][len[1]]=cst[1];

for (int i=2;i<=n;i++){

dp[i][len[i]]=cst[i];

for (it=dp[i-1].begin();it!=dp[i-1].end();it++){

if (dp[i][it->first]==0 || dp[i][it->first]>dp[i-1][it->first])dp[i][it->first]=dp[i-1][it->first];

//警察叔叔，就是上一句犯了SB错！

int lenx=gcd(len[i],it->first);

int val=it->second;

if (dp[i][lenx]==0 || dp[i][lenx]>val+cst[i]) dp[i][lenx]=val+cst[i];

}

int Ans=2147483647;

for (int i=1;i<=n;i++) if (dp[i][1]!=0) Ans=Ans>dp[i][1]?dp[i][1]:Ans;

printf("%dn",Ans);

return 0;

}

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Codeforces Round #290 (Div. 2)

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Codeforces Round #290 (Div. 2)

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