2016河南省第九届ACM程序设计竞赛[正式赛四]

-0:12:10

Overview
Problem
Status
Rank

Current Time:	2016-04-13 13:47:49	Contest Type:	Private
Start Time:	2016-04-13 09:00:00	Contest Status:	Running
End Time:	2016-04-13 14:00:00	Manager:	zhongshijun

	ID	Title
26 / 49	Problem A	A
0 / 11	Problem B	B
3 / 52	Problem C	C
0 / 2	Problem D	D
23 / 158	Problem E	E
0 / 3	Problem F	F
0 / 4	Problem G	G
22 / 60	Problem H	H
25 / 27	Problem I	I

A B C D E F G H I

E - E

Crawling in process... Crawling failed Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

Submit Status

Description

Last week, n students participated in the annual programming contest of Marjar University. Students are labeled from 1 to n. They came to the competition area one by one, one after another in the increasing order of their label. Each of them went in, and before sitting down at his desk, was greeted by his/her friends who were present in the room by shaking hands.

For each student, you are given the number of students who he/she shook hands with when he/she came in the area. For each student, you need to find the maximum number of friends he/she could possibly have. For the sake of simplicity, you just need to print the maximum value of the n numbers described in the previous line.

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains an integer n (1 ≤ n ≤ 100000) -- the number of students. The next line contains n integers a₁, a₂, ..., a_n (0 ≤ a_i < i), where a_i is the number of students who the i-th student shook hands with when he/she came in the area.

Output

For each test case, output an integer denoting the answer.

Sample Input

Sample Output

2
3

#include<stdio.h>
int a[109000];
int main()
{int t,i,n;scanf("%d",&t);while(t--){scanf("%d",&n);int p=0;for(i=1; i<=n; i++){scanf("%d",&a[i]);if(a[i]==0){p++;}}if(p==n){printf("0\n");}else{int sum=0;i=2;while(i<=n){if(a[i]!=0){sum=a[i];for(int j=i+1; j<=n; j++){if(a[j]!=0){if(sum<a[j]){sum=a[j];}else{sum++;}}}printf("%d\n",sum);break;}else{i++;}}}}
}

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