Bus Video System CodeForces - 978E （思维）

The busses in Berland are equipped with a video surveillance system. The system records information about changes in the number of passengers in a bus after stops.

If y−x. So the system records show how number of passengers changed.

The test run was made for single bus and n in chronological order.

Determine the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to w passengers inclusive).

Input

The first line contains two integers (1≤n≤1000,1≤w≤109) — the number of bus stops and the capacity of the bus.

The second line contains a sequence i-th bus stop.

Output

Print the number of possible ways how many people could be in the bus before the first bus stop, if the bus has a capacity equals to 0.

Examples

Input

3 5
2 1 -3

Output

3

Input

2 4
-1 1

Output

4

Input

4 10
2 4 1 2

Output

2

Note

In the first example initially in the bus could be 2 passengers.

In the second example initially in the bus could be 4 passengers.

In the third example initially in the bus could be 1 passenger.

 

题意：

 给你一个含有n个整数的数组，每一个数a[i]代表汽车在站i时，车上增多了a[i]个人，如果a[i]为负，代表减少了人数。

并告诉你这个汽车的最大承载力为w个人，

请你判断初始时汽车上有多少个人，才满足整个数组的情况，。

如果某一个情况，车上的人数为负，或者人数大于w，那么说明这个数组时不合理的，。这时请输出0

思路：

可以抽象为，求这个数组的前缀和数组中的最大值和最小值，。只要最大值不大于容量，再判断下最低值的绝对值不大于容量。就可以说明是合理的。

然后可以的方案数中初始的人数一定是连续的，那么这些人数中的最大值是min(w-maxsum,w) ，即不让过程中容量大于w的最大值。

最小值是max（0,-1*minsum），然后最大值减去最小值+1就是答案了。

 

细节见代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int* p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
// HFUU-QerM
// 21:49:59
ll n;
ll w;
ll a[maxn];
int main()
{
    //freopen("D:\common_text\code_stream\in.txt","r",stdin);
    //freopen("D:\common_text\code_stream\out.txt","w",stdout);
    gbtb;
    cin >> n >> w;
    repd(i, 1, n)
    {
        cin >> a[i];
    }
    ll f = -1e18;
    ll g = 1e18;
    ll v = 0ll;
    repd(i, 1, n)
    {
        v += a[i];
        f = max(f, v);
        g = min(g, v);
    }
    // db(f);
    // db(g);
    if ((abs(f)) > w || abs(g) > w)
    {
        cout << 0 << endl;
    } else
    {
        ll s = w - f;
        ll x = 0ll;
        // db(s);
        s=min(s,w);
        if (g < 0)
        {
            x = -1 * g;
        }
        // db(x);
        if (x > s)
        {
            cout << 0 << endl;
        } else
        {
            cout << s - x + 1ll << endl;
        }
    }




    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}