CF #358 A. Alyona and Numbers
传送门：http://codeforces.com/contest/682/problem/A
题目大意
输入两个数字n,m，分别从1~n,1~m中选择一个数，使它们的和为5的倍数，求方案数。
题解
根据模5意义下直接算方案数
程序
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#include #include #include using namespace std; int n,m; long long Ans=0; int main(){ scanf("%d%d",&n,&m); //0+0 Ans += (long long)(n/5)*(long long)(m/5); //1+4 Ans += (long long)(n/5+(n%5>=1))*(long long)(m/5+(m%5>=4)) + (long long)(n/5+(n%5>=4))*(long long)(m/5+(m%5>=1)); //2+3 Ans += (long long)(n/5+(n%5>=2))*(long long)(m/5+(m%5>=3)) + (long long)(n/5+(n%5>=3))*(long long)(m/5+(m%5>=2)); cout<<Ans<<endl; return 0; } |
原文链接：CF #358 A. Alyona and Numbers
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