Codeforces Round #358 (Div. 2)(A) Alyona and Numbers
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http://codeforces.com/problemset/problem/682/A
题意:
输入n、m,找出所有1<=x<=n,1<=y<=m,使得(x+y)是5的倍数的组合的个数。
代码:
#include <iostream>#include <cstdio>using namespace std;int n,m;long long ans;int main(){ ans=0; scanf("%d %d",&n,&m); for(int i=1;i<=n;i++) { ans+=((m+i)/5-i/5); } printf("%I64d\n",ans); return 0;}
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