CodeForces 131B Opposites Attract

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两重for循环妥妥超时

#include<stdio.h>#include<iostream>#include<math.h>#include<string.h>#include<iomanip>#include<stdlib.h>#include<ctype.h>#include<algorithm>#include<deque>#include<functional>#include<iterator>#include<vector>#include<list>#include<map>#include<queue>#include<set>#include<stack>#define CPY(A, B) memcpy(A, B, sizeof(A))typedef long long LL;typedef unsigned long long uLL;const int MOD = 1e9 + 7;const int INF = 0x3f3f3f3f;const LL INFF = 0x3f3f3f3f3f3f3f3fLL;const double EPS = 1e-9;const double OO = 1e20;const double PI = acos (-1.0);const int dx[] = {-1, 0, 1, 0};const int dy[] = {0, 1, 0, -1};using namespace std;const int maxn=1250;int a[15],b[15],z=0;int main() {    int n,t;    scanf ("%d",&n);    for (int i=0; i<n; ++i) {        scanf ("%d",&t);        if (t>0) {++a[t];}        else if (t<0) {++b[-t];}        else { ++z; }//统计0的个数    }//two for loop will TLE    uLL cnt=0;    for (int i=0; i<11; ++i) {cnt+= (uLL) a[i]*b[i];}//匹配正负    cnt+= (uLL) z* (z-1) /2;//0 and other 0    cout<<cnt<<endl;    return 0;}

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