codeforces 808C

最新推荐文章于 2019-01-16 14:49:00 发布

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本文探讨了如何在满足特定条件下为不同容量的杯子分配固定数量的茶的问题。Polycarp邀请朋友参加茶话会，每位朋友有一个容量不同的杯子，需要确保每个杯子至少装一半的容量且所有茶都要倒完，同时所有朋友都满意。

Tea Party
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Polycarp invited all his friends to the tea party to celebrate the holiday. He has n cups, one for each of his n friends, with volumes a1, a2, …, an. His teapot stores w milliliters of tea (w ≤ a1 + a2 + … + an). Polycarp wants to pour tea in cups in such a way that:

Every cup will contain tea for at least half of its volume
Every cup will contain integer number of milliliters of tea
All the tea from the teapot will be poured into cups
All friends will be satisfied.
Friend with cup i won’t be satisfied, if there exists such cup j that cup i contains less tea than cup j but ai > aj.

For each cup output how many milliliters of tea should be poured in it. If it’s impossible to pour all the tea and satisfy all conditions then output -1.

Input
The first line contains two integer numbers n and w (1 ≤ n ≤ 100, ).

The second line contains n numbers a1, a2, …, an (1 ≤ ai ≤ 100).

Output
Output how many milliliters of tea every cup should contain. If there are multiple answers, print any of them.

If it’s impossible to pour all the tea and satisfy all conditions then output -1.

Examples
input
2 10
8 7
output
6 4
input
4 4
1 1 1 1
output
1 1 1 1
input
3 10
9 8 10
output
-1
Note
In the third example you should pour to the first cup at least 5 milliliters, to the second one at least 4, to the third one at least 5. It sums up to 14, which is greater than 10 milliliters available.
水题，直接上代码，但是CF上第12组数据无法通过，系统自动容错，自己编译器也无法通过
这里写图片描述

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
int main( ){
    int n,w,sum,wa;
    int q,max_num;
    int num1[105];
    int num2[105];
    int sum_num;
    int num3[105];
    scanf("%d%d",&n,&w);
    wa = 0;
    memset(num3, 0, sizeof(num3));
    sum=0;
    q=-1;
    sum_num=0;
    max_num=-1;
    for(int i=1;i<=n;i++){
        scanf("%d",&num1[i]);
        num2[i]=num1[i];
        num3[i]=(num1[i]+1)/2;
        sum+=num1[i];
        sum_num+=num3[i];

    }
    if(sum == w)
    {
        for(int i=1;i<=n;i++)
            printf("%d ",num2[i]);
        return 0;
    }
    if(w<sum_num)
        printf("-1\n");
    else{
        w-=sum_num;
        do{
            num2[q]=-2;
            max_num=-1;
            for(int i=1;i<=n;i++)
                if(max_num<num2[i]){
                    max_num=num2[i];
                    q=i;
                }

            if(w<=(num2[q]-num3[q])){
                num3[q]+=w;
                w=0;

            }
            else
            {
                w-=(num2[q]-num3[q]);
                num3[q]=num1[q];
            }
        }while(w!=0);
        for(int i=1;i<=n;i++)
            printf("%d ",num3[i]);
    }
    return 0;
}