Codeforces 754A - Lesha and array splitting

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概述

One spring day on his way to university Lesha found an array A. Lesha likes to split arrays into several parts. This time Lesha decided to split the array A into several, possibly one, new arrays so that the sum of elements in each of the new arrays is not zero. One more condition is that if we place the new arrays one after another they will form the old array A.

Lesha is tired now so he asked you to split the array. Help Lesha!

Input

The first line contains single integer n (1 ≤ n ≤ 100) — the number of elements in the array A.

The next line contains n integers a₁, a₂, ..., a_n ( - 10³ ≤ a_i ≤ 10³) — the elements of the array A.

Output

If it is not possible to split the array A and satisfy all the constraints, print single line containing "NO" (without quotes).

Otherwise in the first line print "YES" (without quotes). In the next line print single integer k — the number of new arrays. In each of the next k lines print two integers l_i and r_i which denote the subarray A[l_i... r_i] of the initial array Abeing the i-th new array. Integers l_i, r_i should satisfy the following conditions:

l₁ = 1
r_k = n
r_i + 1 = l_i + 1 for each 1 ≤ i < k.

If there are multiple answers, print any of them.

Example

Input

3
1 2 -3

Output

YES
2
1 2
3 3

Input

8
9 -12 3 4 -4 -10 7 3

Output

YES
2
1 2
3 8

Input

1
0

Output

NO

Input

4
1 2 3 -5

Output

YES
4
1 1
2 2
3 3
4 4




题目大意：给你一个数组，现在要你把这个数组分成几部分，要求每部分之和不能等于0

题目分析：如果这个数组不能完成上述要求，说明这个数组的全部元素都等于0，直接输出No。首先遍历一遍。
记录下来有多少个元素不等于0，然后从头开始遍历分组，每个元素，如果不等于0的话，就把他放进这个组里，
自成一组，如果它的后面紧跟着0的话，就把0也放进去跟它一组，贪心求解。
#include <cstdio>
	#include <cstring>
	#include <algorithm>
	#include <iostream>
	using namespace std;
	int a[110],b[110],c[110];
	int main(){
		int k,m,n;
		int cnt;
		scanf("%d",&n);
		cnt=m=0;
		for(int i=1 ;i<=n;i++){
			scanf("%d",&a[i]);
			if(a[i]!=0)
			  m++;
		}
		if(m == 0){
			printf("NOn");
		}
		else {	
			for(int i=1;i<=n;i++){
			b[++cnt] = i;
			while(a[i] == 0)
			   i++;
			if(--m == 0)
	            break;
			c[cnt]= i;
			}
			c[cnt]=n;
			printf("YESn%dn",cnt);
			for(int i=1;i<=cnt;i++){
				printf("%d %dn",b[i],c[i]);
			}
		}
		return 0;
	}