2018.5.12—记录菜菜的一场比赛_you are given an array a consisting of n distinct

写个博客记录一下菜鸟成长记~

赛后认真总结反思了一下，还是基础不牢固，思路有了要么是细节处理有问题，要么是题意有遗漏。

A:说是规律。。反正我没想到。。还在研究。

B:

You are given an array a consisting of n elements a1, a2, ..., an. Array a has a special property, which is:

ai = ( ai - 1 + 1) % m, for each i (1 < i ≤ n)
You are given the array a with some lost elements from it, each lost element is replaced by -1. Your task is to find all the lost elements again, can you?
Input
The first line contains an integer T, where T is the number of test cases.
The first line of each test case contains two integers n and m (1 ≤ n ≤ 1000) (1 ≤ m ≤ 109), where n is the size of the array, and m is the described modulus in the problem statement.
The second line of each test case contains n integers a1, a2, ..., an ( - 1 ≤ ai < m), giving the array a. If the ith element is lost, then ai will be -1. Otherwise, ai will be a non-negative integer less than m.
It is guaranteed that the input is correct, and there is at least one non-lost element in the given array.
Output
For each test case, print a single line containing n integers a1, a2, ..., an, giving the array a after finding all the lost elements.
It is guaranteed that an answer exists for the given input.
Example
Input
4
5 10
1 2 3 4 5
4 10
7 -1 9 -1
6 7
5 -1 -1 1 2 3
6 10
5 -1 7 -1 9 0
Output
1 2 3 4 5
7 8 9 0
5 6 0 1 2 3

5 6 7 8 9 0

水题：就是根据一个已知的就能推其它的。

代码：

#include<bits/stdc++.h>
using namespace std;
int a[1010];
int main()
{
    int t;
    int n,m;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        int lala;
        for(int i=1;i<=n;i++)
        {
            if(a[i]!=-1)
            {
                a[i+1]=(a[i]+1)%m;
                lala=i;
            for(int j=i+2;j<=n;j++)
            {
                a[j]=(a[j-1]+1)%m;
            }
            break;
            }
        }
        for(int i=lala-1;i>=1;i--)
        {
            a[i]=a[i+1]-1;
            if(a[i]<0)
                a[i]+=m;
        }
        for(int i=1;i<=n;i++)
        {
            cout<<a[i]<<" ";
        }
        cout<<endl;
    }
    return 0;
}

C:二分！！！

You are given an array a consisting of n element a1, a2, ..., an. For each element ai you must find another element aj (i ≠ j), such that the summation of ai and aj mod (109 + 7) (i.e. (ai + aj) % (109 + 7)) is as maximal as possible.


Can you help judges by solving this hard problem?


Input
The first line contains an integer T, where T is the number of test cases.


The first line of each test contains an integer n (2 ≤ n ≤ 10