NEFU 2016省赛演练一 I题 (模拟题)

这题没名字

Problem:I

Time Limit:2000ms

Memory Limit:65535K

Description

Now give you an interger m and a sequence: s[1], s[2], ...... , s[N] in not decreasing order. The length of the sequence is n. Your task is to find out how many pair of a and b satisfy a+b=m. Note that each element of the sequence can use once at most. 

Input

The input consists of multiple test cases. The first line of each test case contains three integers N（0<n<=1000000）, M(0<m<=1000000000), which denote the length of the sequence and the sum of a+b, respectively. The next line give n interger（0<s[i]<=10000000）。 Note the input is huge.

Output

For each case, output the num of pairs in one line.

Sample Input

3 3
1 1 2
3 2
1 1 1
5 5
1 1 2 3 4

Sample Output

1
1
2

题意：给定n和m，输入n个数，判断在这些数中有多少组两个数的和为m。
题解：用while跑是最快的，跑n存在风险，跑数组可能超时，跑m一定超时。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
typedef long long ll;
ll a[];
int main()
{
    ll i,n,m,data;
    while(scanf("%lld%lld",&n,&m)!=EOF)
    {
        memset(a,,sizeof(a));
        for(i=;i<n;i++)
        scanf("%lld",&a[i]);
        ll ans=;
        sort(a,a+n);
        ll l=,r=n-;
        while(l<r)
        {
            if(a[l]+a[r]>m)
            r--;
            else if(a[l]+a[r]<m)
            l++;
            else
            {
                ans++;
                r--;
                l++;
            }
        }
        printf("%lld\n",ans);
    }
    return ;
}