【UVa1152】4 Values whose Sum is 0 模拟
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原题:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3593
题意:
给出四个集合abcd,求有多少组解。n<=4000。
直接四层循环会超时,所以将等式转换成a[i]+b[i]=-(c[i]+d[i]),预处理一下a[i]+b[i]就可以把复杂度降到O(n^2*logn)。(用upper_bound-lower_bound可以求出相等的个数)
#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>#define lim 4010using namespace std;int n,T;int a[lim],b[lim],c[lim],d[lim],res[lim*lim];int main(){scanf("%d",&T);while(T--){int tot=0,ans=0;scanf("%d",&n);for (int i=0;i<n;i++)scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);for (int i=0;i<n;i++)for (int j=0;j<n;j++) res[tot++]=a[i]+b[j];//先预处理a[i]+b[i]sort(res,res+tot);for (int i=0;i<n;i++)for (int j=0;j<n;j++) ans+=upper_bound(res,res+tot,-(c[i]+d[j]))-lower_bound(res,res+tot,-(c[i]+d[j]));//这样就求出了相等的个数printf("%d\n",ans);if (T) printf("\n");//不加会PE QwQ}return 0;}
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