Lesha and array splitting
time limit per test 2 second memory limit per test 256 megabytes
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.

传送门:CF754A
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,...,an (-10^3≤ai≤10^3) — 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 li and ri which denote the subarray A[li... ri] of the initial array A being the i-th new array. Integers li, ri should satisfy the following conditions: l1=1 rk=n ri+1=li+1 for each 1≤i<k. If there are multiple answers, print any of them.

Sample Input

1
2
3
1 2 -3

Sample Output

1
2
3
4
YES
2
1 2
3 3

题解

trick题,要能分成>=1个的所有和不为零的子串 其实只要判断所有和为不为零 不为零 那就是全部分成一组 为零 从前往后找第一个和不为零的下标,这样就能分成两串 如果找不到 说明不能拆分 但我那时候没这么做,= = 我是从前往后扫,每次sum+当前值,如果sum!=0就分开,然后sum=0,继续往下走 如果最后sum还等于0,看它前面有没有分过,分过的话就并到前面去。
## AC code:(不包含输入类)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
//package A;  
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
while(sc.hasNext()){
int n=sc.nextInt();
int[]shu=new int[n+1];
for(int i=1;i<=n;i++){
shu[i]=sc.nextInt();
}
int sum=0;
ArrayList<Integer>list=new ArrayList<Integer>();
for(int i=1;i<=n;i++){
sum=sum+shu[i];
if(sum!=0){
list.add(i);
sum=0;
}
}
if(sum==0){
if(list.size()>=1){
list.set(list.size()-1,n);
}
}
if(list.size()>0){
System.out.println("YES");
System.out.println(list.size());
System.out.print("1 ");
System.out.println(list.get(0));
for(int i=1;i<list.size();i++){
System.out.println(list.get(i-1)+1+" "+list.get(i));
}
}
else{
System.out.println("NO");
}
}
}
}