Educational Codeforces Round 21 D. Array Division(前缀和，二分)

D. Array Division
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Vasya has an array a consisting of positive integer numbers. Vasya wants to divide this array into two non-empty consecutive parts (the prefix and the suffix) so that the sum of all elements in the first part equals to the sum of elements in the second part. It is not always possible, so Vasya will move some element before dividing the array (Vasya will erase some element and insert it into an arbitrary position).

Inserting an element in the same position he was erased from is also considered moving.

Can Vasya divide the array after choosing the right element to move and its new position?

Input
The first line contains single integer n (1 ≤ n ≤ 100000) — the size of the array.

The second line contains n integers a1, a2… an (1 ≤ ai ≤ 109) — the elements of the array.

Output
Print YES if Vasya can divide the array after moving one element. Otherwise print NO.

Examples
input
3
1 3 2
output
YES
input
5
1 2 3 4 5
output
NO
input
5
2 2 3 4 5
output
YES
Note
In the first example Vasya can move the second element to the end of the array.

In the second example no move can make the division possible.

In the third example Vasya can move the fourth element by one position to the left.

题意是给出一个数组，求能否移动一个数，使得某个位置前的所有数的和是总和的一半。

输入预处理下，求出前缀和，因为所有的数都是整数且互不相同，所以前缀和肯定是递增的。
对于任意一个数，如果将它放在某个数的后面使得条件满足，那么就有 sum[x] = sum/2 - b[i]; ( x>i)
如果将它放在某个数的前面使得条件满足，那么同上就有 sum[x] = sum/2 + b[i]; (x < i);

那么问题就变成了求x了，二分下标

做题的时候只想到了放在某个数的后面，没有考虑前面的情况，真是菜啊

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#include<string.h>
#include<algorithm>

using namespace std;

long long sum[100010];
int b[100010];

int main()
{
    int n;

    scanf("%d",&n);

    for(int i=0;i<n;i++)
        {
            int tmp;
            scanf("%d",&tmp);

            b[i] = tmp;

            if(i==0)  sum[i] = tmp;
            else    sum[i] += sum[i-1] + tmp;
        }

    if(sum[n-1] % 2 == 1)  // the sum is odd
        {
            printf("NO\n");
            return 0;
        }

    long long tmp = sum[n-1] / 2;

    int i;
    for( i=0;i<n;i++)
        {   
            if(sum[i] == tmp)
                {
                    printf("YES\n");
                    return 0;
                }

            int l=i,r=n-1;
            int mid;
            long long f = tmp + b[i];
            while(l <= r)
            {
                mid = (l+r)/2;
                if( sum[mid] > f )
                    r = mid-1;  
                else if(sum[mid] < f)
                    l = mid+1;
                else 
                    {
                        if(mid > i)
                            {
                                printf("YES\n");
                                return 0;
                            }
                        else break; 
                    } 
            }

            f = tmp - b[i];
            l=0;r=n-1;
            while(l<=r)
            {
                mid = (l+r)/2;
                if(sum[mid] >f )
                    r = mid-1;
                else if(sum[mid] < f)
                    l = mid+1;
                else 
                    {
                        if(mid<i)
                        {
                            printf("YES\n");
                            return 0;
                        }
                        else break;
                    }
             } 
        }
        if(i==n)
            printf("NO\n");

    return 0;
}