2 条题解
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0
#include<bits/stdc++.h> using namespace std; typedef long long LL; LL gcd(LL x, LL y) {if(x==0) return y; else return gcd(y%x, x);} int main(){ //freopen("a.in", "r", stdin); int T; scanf("%d", &T); while(T--){ LL n, m, ans=0; scanf("%lld%lld", &n, &m); for(LL i=1; i<=n/i; i++) for(LL j=1; j<=m/j; j++) if(gcd(i, j)==1) ans+=min(n/i, m/j)/(i+j); printf("%lld\n", ans); } return 0; } -
0
#include<bits/stdc++.h> using namespace std; typedef long long LL; LL gcd(LL x, LL y) {if(x==0) return y; else return gcd(y%x, x);} int main(){ //freopen("a.in", "r", stdin); int T; scanf("%d", &T); while(T--){ LL n, m, ans=0; scanf("%lld%lld", &n, &m); for(LL i=1; i<=n/i; i++) for(LL j=1; j<=m/j; j++) if(gcd(i, j)==1) ans+=min(n/i, m/j)/(i+j); printf("%lld\n", ans); } return 0; }
- 1
信息
- ID
- 2118
- 时间
- 1500ms
- 内存
- 256MiB
- 难度
- 8
- 标签
- 递交数
- 22
- 已通过
- 5
- 上传者