Codeforces Round #257 (Div. 2) A题

A. Jzzhu and Children

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There are n children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from 1to n. The i-th child wants to get at least ai candies.

Jzzhu asks children to line up. Initially, the i-th child stands at the i-th place of the line. Then Jzzhu start distribution of the candies. He follows the algorithm:

1. Give m candies to the first child of the line.
2. If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home.
3. Repeat the first two steps while the line is not empty.

Consider all the children in the order they go home. Jzzhu wants to know, which child will be the last in this order?

Input

The first line contains two integers n, m (1 ≤ n ≤ 100; 1 ≤ m ≤ 100). The second line contains n integersa1, a2, ..., an (1 ≤ ai ≤ 100).

Output

Output a single integer, representing the number of the last child.

Sample test(s)

input

5 2
1 3 1 4 2

output

4

input

6 4
1 1 2 2 3 3

output

6

Note

Let's consider the first sample.

Firstly child 1 gets 2 candies and go home. Then child 2 gets 2 candies and go to the end of the line. Currently the line looks like [3, 4, 5, 2] (indices of the children in order of the line). Then child 3 gets 2 candies and go home, and then child 4 gets 2 candies and goes to the end of the line. Currently the line looks like [5, 2, 4]. Then child 5 gets 2 candies and goes home. Then child 2 gets two candies and goes home, and finally child 4 gets 2 candies and goes home.

Child 4 is the last one who goes home.

---------------------------------------------------

题目意思是给几个孩子发糖，每个孩子都有自己要的糖的个数，但是每次只是固定的发一定量的糖，每个没拿到自己期望的糖数就到队尾继续排队，问你最后走的孩子是几号

、、、、、、、、、、、、、、、、、、、、、、、、、、

 #include <stdio.h>
 #include <string.h>
 #include <stdlib.h>
 #include <iostream>
 #include <algorithm>

 using namespace std;

 int main()
 {
     int n,m,i,j;
     while(scanf("%d%d",&n,&m)!=EOF)
     {
         int str[],maxx=,st;
         for(i=;i<n;i++)
         {
             scanf("%d",&str[i]);
             if(maxx<str[i])
             {
                 maxx=str[i];
             }
         }
         st=maxx/m;
         if(st*m<maxx)
         {
             st++;
         }//printf("%d%d**",st,maxx);
         for(i=;i<n;i++)
         {
             str[i]-=(st-)*m;//printf("%d**",str[i]);
         }
         int res=,cas;
         for(i=;i<n;i++)
         {
             if(str[i]>)
             {
                 //res=str[i];
                 cas=i;
             }
         }
         printf("%d\n",cas+);
     }
     return ;
 }

这是队列模拟的

 #include <stdio.h>
 #include <iostream>
 #include <algorithm>
 #include <queue>

 using namespace std;

 typedef struct stu
 {
     int id,m;
 }stu;
 stu str[];

 int main()
 {
     int i,j,n,m,k,t;

     while(scanf("%d%d",&n,&m)!=EOF)
     {
         queue<stu >q;

         for(i=;i<n;i++)
         {
             scanf("%d",&str[i].m);
             str[i].id=i+;
             q.push(str[i]);
         }
         int tmp=;
         while(!q.empty())
         {
             stu x=q.front();
             q.pop();
             if(x.m<=m)
             {
                 tmp=x.id;
             }
             else
             {
                 x.m-=m;
                 q.push(x);
             }
         }
         printf("%d\n",tmp);
     }
     return ;
 }

这题A题卡了我好久，主要因为数组模拟没有想到找最大的那个趟数来模拟

经验不足……