Codeforces Round #512 (Div. 2, based on Technocup 2019 Elimination Round 1) C. Vasya and Golden Ticket 【。。。】

任意门：http://codeforces.com/contest/1058/problem/C

C. Vasya and Golden Ticket

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Recently Vasya found a golden ticket — a sequence which consists of $\mathrm{nn digits a1a2\dots ana1a2\dots an.\; Vasya\; considers\; a\; ticket\; to\; be\; lucky\; if\; it\; can\; be\; divided\; into\; two\; or\; more\; non-intersecting\; segments\; with\; equal\; sums.\; For\; example,\; ticket 350178350178 is\; lucky\; since\; it\; can\; be\; divided\; into\; three\; segments 350350, 1717 and 88: 3+5+0=1+7=83+5+0=1+7=8.\; Note\; that\; each\; digit\; of\; sequence\; should\; belong\; to exactly one\; segment.}$

Help Vasya! Tell him if the golden ticket he found is lucky or not.

Input

The first line contains one integer $\mathrm{nn (2\le n\le 1002\le n\le 100)\; —\; the\; number\; of\; digits\; in\; the\; ticket.}$

The second line contains $\mathrm{nn digits a1a2\dots ana1a2\dots an (0\le ai\le 90\le ai\le 9)\; —\; the\; golden\; ticket.\; Digits\; are\; printed\; without\; spaces.}$

Output

If the golden ticket is lucky then print "YES", otherwise print "NO" (both case insensitive).

Examples

input

Copy

5
73452

output

Copy

YES

input

Copy

4
1248

output

Copy

NO

Note

In the first example the ticket can be divided into $\mathrm{77, 3434 and 5252: 7=3+4=5+27=3+4=5+2.}$

In the second example it is impossible to divide ticket into segments with equal sum.

题意概括：

一串长度为N的数字，判断是否能分成相同和的若干块。

解题思路：

记一道被自己蠢哭的水题，主要是理解题意的问题；

一开始以为分块可以不连续，又是排序又乱七八糟搞一大堆；

结果分块是连续的，活生生的贪心水题。

读题很重要。

AC code：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
const int MAXN = 102;
char str[MAXN];
int N, sum;
int main()
{
    scanf("%d", &N);
    scanf("%s", &str);
    bool flag = true;
    for(int i = 0; i < N-1; i++){
        sum += str[i]-'0';
        int pos = i+1;
        flag = true;
        while(pos < N){
            int sum2 = str[pos++]-'0';
            while(pos < N && sum2+str[pos]-'0' <= sum){
                sum2+=str[pos++]-'0';
            }
            if(sum2 != sum) {flag = false; break;}
        }
        if(flag) {puts("YES"); break;}
    }
    if(flag == false) puts("NO");
    return 0;
}