fzu 2058 问题转化 + 二分

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题意

A array contain N number, can you tell me how many different pair i,j that satisfy a[i] + a[j] = M and 1<=i < j <=N.

0 < N <= 100000, 0 <= M <= 1000000000, -1000000000 <= a[i] <= 1000000000

思路：

用map记录会超时

所以想到手写二分。思考一下其实满足题目条件的对数和每个数字的位置并没有关系——排序+二分查找（范围来确定个数）即可。

code：

#include<cstdio>#include<iostream>#include<sstream>#include<cstring>#include<algorithm>#include<vector>#include<string>#include<queue>#include<map>#include<set>#include<cmath>#include<cctype>#include<cstdlib>using namespace std;#define INF 0x3f3f3f3f#define PI acos(-1.0)#define mem(a, b) memset(a, b, sizeof(a))#define mod 1000000007typedef pair<int,int> pii;typedef long long LL;//------------------------------const int maxn = 200005;int a[maxn];int n,m;int Scan(){    int f=1, ret=0;    char c;    while((c=getchar())==' ' || c=='\n') ;    if(c=='-') f=-1;    else if(c!='+') ret+=c-'0';    while((c=getchar())<='9' && c>='0')        ret=ret*10+c-'0';    return ret*f;}int main(){    while(scanf("%d%d",&n,&m) != EOF){        for(int i = 0; i < n; i++){            a[i] = Scan();        }        sort(a,a+n);        long long cnt = 0;        for(int i = 0; i < n; i++){            int id1 = lower_bound(a+i+1, a+n, m-a[i]) - a;            int id2 = upper_bound(a+i+1, a+n, m-a[i]) - a;            cnt += (id2 - id1);        }        cout << cnt << endl;    }    return 0;}

一TLE吓得我都开挂了....T_T

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