Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) - C

题目链接：http://codeforces.com/contest/831/problem/C

题意：给定k个评委，n个中间结果。 假设参赛者初始分数为x，按顺序累加这k个评委的给分后得到k个结果，在这个过程中，某人只记得其中n次结果(n次结果不一样按照时间顺序来排列)，问你参赛者的初始合法分数有多少种。

思路：维护一个评委给分的前缀和，然后枚举每一种可能的初始分数，然后判断当前枚举的这个分数经过k次累加分数后得到的k个结果有n个结果等于给定的中间结果即可。

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<queue>
#include<vector>
#include<time.h>
#include<stack>
#include<set>
#include<cmath>
#include<functional>
#include<cstdlib>
using namespace std;
typedef long long int LL;
typedef unsigned long long int ULL;
const LL INF = 9223372036854775807L;
const int inf = 0x3f3f3f3f;
const int MAXN =  + ;
const int MAXB = 8e6 + ;
set<LL>se;
LL val[MAXN], source[MAXN],pre[MAXN];
LL statu[MAXN];
bool visb[MAXB],vis[MAXB];
int main(){
#ifdef kirito
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    int start = clock();
    int n, k;
    while (~scanf("%d%d",&k,&n)){
        se.clear();
        memset(visb, , sizeof(visb));
        memset(pre, , sizeof(pre));
        for (int i = ; i <= k; i++){
            scanf("%I64d", &val[i]);
            pre[i] += pre[i - ] + val[i];
        }
        for (int i = ; i <= n; i++){
            scanf("%I64d", &source[i]);
            visb[source[i] + ] = ;
        }
        for (int i = ; i <= k; i++){
            int cnt = ;
            statu[] = source[]-pre[i];
            memset(vis, , sizeof(vis));
            if (se.count(statu[])){
                continue;
            }
            for (int j = ; j <= k; j++){
                statu[j] = statu[j-] + val[j];
                if (statu[j] + <MAXB&&visb[statu[j] + ] && vis[statu[j] + ] == false){
                    cnt++;
                    vis[statu[j] + ] = true;
                }
            }
            if (cnt == n){
                se.insert(statu[]);
            }
        }
        printf("%d\n", se.size());
    }
#ifdef LOCAL_TIME
    cout << "[Finished in " << clock() - start << " ms]" << endl;
#endif
    return ;
}