nyoj 791 Color the fence(贪心)
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Color the fence
时间限制:1000 ms | 内存限制:65535 KB
难度:2
- 描述
Tom has fallen in love with Mary. Now Tom wants to show his love and write a number on the fence opposite to
Mary’s house. Tom thinks that the larger the numbers is, the more chance to win Mary’s heart he has.
Unfortunately, Tom could only get V liters paint. He did the math and concluded that digit i requires ai liters paint.
Besides,Tom heard that Mary doesn’t like zero.That’s why Tom won’t use them in his number.
Help Tom find the maximum number he can write on the fence.
- 输入
- There are multiple test cases.
Each case the first line contains a nonnegative integer V(0≤V≤10^6).
The second line contains nine positive integers a1,a2,……,a9(1≤ai≤10^5). - 输出
- Printf the maximum number Tom can write on the fence. If he has too little paint for any digit, print -1.
- 样例输入
55 4 3 2 1 2 3 4 529 11 1 12 5 8 9 10 6
- 样例输出
5555533
解题思路:
首先要肯定一点,数字位数越多,肯定会越大。所以这里可以采用贪心的思想,假设最小的颜料的用量为min,那么最长的位数可以达到V / min,接下来我们就枚举每一位可能的数字即可。这里还需要考虑,取了某一个数后,剩下的是否还可以保证最大。。
#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int a[10];int main(){ int n; while(~scanf("%d", &n)) { int Min = 0x3f3f3f3f; for(int i = 1; i < 10; i++) { scanf("%d", &a[i]); Min = min(Min, a[i]); } if(n < Min) { puts("-1"); continue; } for(int i = n/Min-1; i >= 0; i--) { for(int j = 9; j > 0; j--) { if(n >= a[j] && (n - a[j])/Min >= i) //保证去了j之,还可以最长。。 { printf("%d", j); n -= a[j]; break; } } } puts(""); } return 0;}
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