Codeforces Round #540 Tanya and Candies 预处理

http://codeforces.com/contest/1118/problem/B

题目大意，给你一个序列，删去一个数值之后，要求剩下序列奇数和偶数的和相同，问有多少种删法。

思路：预处理奇数和偶数和即可

（算法能力康复性训练中......）

//看看会不会爆int!数组会不会少了一维！

//取物问题一定要小心先手胜利的条件

#include <bits/stdc++.h>

#pragma comment(linker,"/STACK:102400000,102400000")

#define LL long long

#define ALL(a) a.begin(), a.end()

#define pb push_back

#define mk make_pair

#define fi first

#define se second

#define haha printf("haha\n")

using namespace std;

const int maxn = 2e5 + ;

int a[maxn];

//表示到当前的偶数或奇数和

int odd1[maxn], even1[maxn];//前往后

int n;

int main(){

    cin >> n;

    for (int i = ; i <= n ;i++){

        scanf("%d", a + i);

    }

    int sum = ;

    for (int i = ; i <= n; i++){

        even1[i] = even1[i-];

        odd1[i] = odd1[i-];

        sum += a[i];

        if (i %  == ) odd1[i] += a[i];

        else even1[i] += a[i];

    }

    int ans = ;

    for (int i = ; i <= n; i++){

        if (i %  == ){

            int odd = even1[n] - even1[i] + odd1[i] - a[i];

            int even = sum - a[i] - odd;

            if (odd == even) ans++;

        }

        else {

            int odd = odd1[i] + even1[n] - even1[i];

            int even = sum - odd - a[i];

            if (odd == even) ans++;

        }

    }

    printf("%d\n", ans);

    return ;

}