B - Array Reconstructing

题面：

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.

从最开始一步步往后推，要先找到第一个不为-1的数，以它为起点向两侧推过去，如果后一个数为0，那么前一个数就为m-1是一个要考虑到的小情况，然后没什么坑了，暴力模拟。

#include<stdio.h>
int main()
{
    int t;
    long long int n,m;
    scanf("%d",&t);
    long long int a[1100];
    while(t--)
    {
        scanf("%lld%lld",&n,&m);
        for(int i=0; i<n; i++)
        {
            scanf("%lld",&a[i]);
        }
        int flag;
        for(int i=0; i<n; i++)
        {
             if(a[i]!=-1)
             {
                 flag = i;
                 break;
             }
        }
         for(int i=flag-1; i>=0; i--)
         {
             if(a[i+1]==0)
                a[i] = m-1;
             else
                a[i] = a[i+1]-1;
         }
         for(int i=flag+1; i<n; i++)
         {
             a[i] = (a[i-1]+1)%m;
         }
         for(int i = 0;i<n;i++)
         {
             if(i==0)
                printf("%lld",a[i]);
             else
                printf(" %lld",a[i]);
         }
         printf("\n");
    }
    return 0;
}