Codeforces Round #202 (Div. 2) B. Color the Fence

最新推荐文章于 2020-03-23 15:48:09 发布

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Igor计划在Tanya家对面的围栏上用油漆写下一个数字来表达爱意，但他的油漆有限。此算法帮助Igor找到使用现有油漆能写出的最大数字，同时避免使用Tanya不喜欢的零。通过贪心策略，找到最小的油漆消耗，然后尽可能多地重复使用该数字。

B. Color the Fence

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Igor has fallen in love with Tanya. Now Igor wants to show his feelings and write a number on the fence opposite to Tanya's house. Igor thinks that the larger the number is, the more chance to win Tanya's heart he has.
Unfortunately, Igor could only get v liters of paint. He did the math and concluded that digit d requires ad liters of paint. Besides, Igor heard that Tanya doesn't like zeroes. That's why Igor won't use them in his number.
Help Igor find the maximum number he can write on the fence.
Input
The first line contains a positive integer v (0 ≤ v ≤ 106). The second line contains nine positive integers a1, a2, ..., a9 (1 ≤ ai ≤ 105).
Output
Print the maximum number Igor can write on the fence. If he has too little paint for any digit (so, he cannot write anything), print -1.
Examples
Input
Copy
5
5 4 3 2 1 2 3 4 5
Output
Copy
55555
Input
Copy
2
9 11 1 12 5 8 9 10 6
Output
Copy
33
Input
Copy
0
1 1 1 1 1 1 1 1 1
Output
Copy
-1

思路：

贪心，找到最小权值，然后暴力尝试替换

代码：

#include<bits/stdc++.h>
using namespace std;
int num[10];
int main()
{
    int v,mmin = 1e7;
    scanf("%d",&v);
    for (int i = 1;i < 10;i ++)
    {
        scanf("%d",&num[i]);
        if (num[i] <= mmin) mmin = num[i];
    }
    if (v < mmin)
    {
        puts("-1\n");
        return 0;
    }
    int t = v / mmin;
    while (t --)
        for (int i = 9;i >= 1;i --)
        {
            if (v >= num[i] && (v - num[i]) / mmin >= t)
            {
                printf("%d",i);
                v -= num[i];
                break;
            }
        }
    putchar('\n');
    return 0;
}