Handshakes(思维) 2016（暴力）

Handshakes

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

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
题解：水题，就是参加运动会，依次进来n个人，每来一个都有朋友给他握手，问最多有多少个朋友，有两种可能；
一：只要有人给其握手，那么把大于等于1的数字个数加起来，相当于第一个人进来给他们都握手。
二：当前进来的人给其握手的最大值；
比较第一和第二种情况的最大值就好了；
代码：

#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

int main()

{

    int t;

    scanf("%d",&t);

    while(t--)

    {

        long long a,maxx=,n;

        scanf("%lld",&n);

        for(int i=;i<n;i++)

        {

            scanf("%lld",&a);

            if(a>)

                maxx++;

            maxx=max(maxx,a);

        }

        printf("%lld\n",maxx);

    }

    return ;

}

2016

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

Submit Status Practice ZOJ 3935

Description

In mathematics, a polygonal number is a number represented as dots or pebbles arranged in the shape of a regular polygon. The dots are thought of as alphas (units). These are one type of 2-dimensional figurate numbers. The following picture shows how triangular numbers, square numbers, pentagonal numbers and hexagonal numbers represented as dots arranged in the shape of corresponding regular polygon.

2016 is not only a leap year but also a triangular and hexagonal year. If you are patient enough, you can count the number of the dots in the left triangle or in the right hexagon in the following picture. The number of dots in each shape is 2016.

Therefore, 2016 is a triangular-hexagonal-leap year. The previous triangular-hexagonal-leap year is 1540 and the next is 2556. So living to see 2016 is very rare experience.

You task is to list the triangular-hexagonal-leap years from 2016 to 990528. 990528 is also a triangular-hexagonal-leap year.

Input

This problem has no input.

Output

Please print each triangular-hexagonal-leap year in increasing order.

For example, if you are asked to list the triangular-hexagonal-leap years from 780 to 2556, the output should be:

Sample Output

2016

2556

...  <-- some lines are skipped

990528
题解：就是找2016--990528之间的闰年，满足三角形以及六边形的那两个公式的年份；
n(n + 1)/2, n(2n - 1);
代码：

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<string>

#include<cstdlib>

using namespace std;

const int INF = 0x3f3f3f3f;

bool triang(int x){

    int t = (int)sqrt(1.0 +  * x);

    if(t * t !=  +  * x)

        return false;

    if((t - ) %  != )

        return false;

//    if(x == 2016)printf("%d %d %d\n",t, t * t, 1 + 8 * x);

    return true;

}

bool hex(int x){

    int t = (int)sqrt(1.0 +  * x);

//    if(x == 2016)printf("%d %d %d\n",t, t * t, 1 + 8 * x);

    if(t * t !=  +  * x)

        return false;

    if((t + ) %  != )

        return false;

    return true;

}

bool leap(int x){

    if(x %  ==  || (x %  ==  && x %  != ))

        return true;

    else

        return false;

}

int main(){

    for(int i = ; i <= ; i++){

        if(leap(i) && triang(i) && hex(i))

            printf("%d\n", i);

    }

    return ;

}