cf(415 A,B)

A. Mashmokh and Lights
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Mashmokh works in a factory. At the end of each day he must turn off all of the lights.

The lights on the factory are indexed from 1 to n. There are n buttons in Mashmokh's room indexed from 1 to n as well. If Mashmokh pushes button with index i, then each light with index not less than i that is still turned on turns off.

Mashmokh is not very clever. So instead of pushing the first button he pushes some of the buttons randomly each night. He pushed mdistinct buttons b1, b2, ..., bm (the buttons were pushed consecutively in the given order) this night. Now he wants to know for each light the index of the button that turned this light off. Please note that the index of button bi is actually bi, not i.

Please, help Mashmokh, print these indices.

Input

The first line of the input contains two space-separated integers n and m (1 ≤ n, m ≤ 100), the number of the factory lights and the pushed buttons respectively. The next line contains m distinct space-separated integers b1, b2, ..., bm (1 ≤ bi ≤ n).

It is guaranteed that all lights will be turned off after pushing all buttons.

Output

Output n space-separated integers where the i-th number is index of the button that turns the i-th light off.

Sample test(s)
input
5 4
4 3 1 2
output
1 1 3 4 4 
input
5 5
5 4 3 2 1
output
1 2 3 4 5 
Note

In the first sample, after pressing button number 4, lights 4 and 5 are turned off and lights 1, 2 and 3 are still on. Then after pressing button number 3, light number 3 is turned off as well. Pressing button number 1 turns off lights number 1 and 2 as well so pressing button number 2 in the end has no effect. Thus button number 4 turned lights 4 and 5 off, button number 3 turned light 3 off and button number 1 turned light 1 and 2 off.

这套题真是考验英语水平。。。。

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
typedef long long ll;
int a[105],b[105]={0};
int main()
{
    int n,m;
    cin>>n>>m;
    for(int i=0;i<m;i++)
    {
        cin>>a[i];
        for(int j=a[i];j<=n;j++)
            if(b[j]==0)
            b[j]=a[i];
            else break;
    }
    for(int i=1;i<=n;i++)
        cout<<b[i]<<' ';
        cout<<endl;
    return 0;
}


B. Mashmokh and Tokens
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Bimokh is Mashmokh's boss. For the following n days he decided to pay to his workers in a new way. At the beginning of each day he will give each worker a certain amount of tokens. Then at the end of each day each worker can give some of his tokens back to get a certain amount of money. The worker can save the rest of tokens but he can't use it in any other day to get more money. If a worker gives backw tokens then he'll get  dollars.

Mashmokh likes the tokens however he likes money more. That's why he wants to save as many tokens as possible so that the amount of money he gets is maximal possible each day. He has n numbers x1, x2, ..., xn. Number xi is the number of tokens given to each worker on the i-th day. Help him calculate for each of n days the number of tokens he can save.

Input

The first line of input contains three space-separated integers n, a, b (1 ≤ n ≤ 105; 1 ≤ a, b ≤ 109). The second line of input contains nspace-separated integers x1, x2, ..., xn (1 ≤ xi ≤ 109).

Output

Output n space-separated integers. The i-th of them is the number of tokens Mashmokh can save on the i-th day.

Sample test(s)
input
5 1 4
12 6 11 9 1
output
0 2 3 1 1 
input
3 1 2
1 2 3
output
1 0 1 
input
1 1 1
1
output
0 


读半天题目都没读懂真是醉了。。。。。。。说一下题意:每天得到ai个可以转换钱的东西(可以比喻成银行卡啥的),只不过这个东西只能当天去换钱,不能每天攒到一起了凑整别的天再去换,换得[w*a/b]块钱,方括号表示取整,可是这个人虽然喜欢这个卡,但是毕竟更喜欢money呀!!所以呢,在换得最大钱数的情况下,有一些卡是可以留下来积攒的,所以让我们求每天最多可以积攒多少这种卡?很简单吧!可是读题是硬伤T_T……

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
typedef long long ll;

ll x;

int main()
{
    int n;ll a,b;
    cin>>n>>a>>b;
    for(int i=0;i<n;i++)
    {
        cin>>x;
        ll tem=x*a/b;
        //cout<<tem;
        ll t=tem*b,tt=x*a-t;
        cout<<tt/a<<' ';
    }

    return 0;
}


posted @ 2015-11-17 18:00  martinue  阅读(167)  评论(0编辑  收藏  举报