贪心,设最小的消耗为minn,最多可以有cnt = v / minn 个数字。 对于每一位数,根据贪心思想,从9->1循环。在每一个循环,检查剩余的 v 是否要比当前数字的消耗量a[ j ]大)(v-a[j] >= 0),此外还要检查消耗了a[ j ]后,是否还能够 cnt 个数((v-a[j])/minn >= i-1)。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <cmath>
using namespace std;
const int INF = 0x7fffffff;
int main()
{
int v;
while(cin >> v)
{
int a[10];
int minn = INF;
int index = 1;
for(int i = 1; i <= 9; ++i)
{
cin >> a[i];
if(a[i] <= minn)
{
minn = a[i];
index = i;
}
}
if(v < minn)
{
cout << -1 << endl;
continue;
}
int cnt = v/minn;
for(int i = cnt; i > 0; --i)
{
for(int j = 9; j > 0; --j)
{
if(v-a[j] >= 0 && (v-a[j])/minn >= i-1)
{
v -= a[j];
cout << j;
break;
}
}
}
cout << endl;
}
return 0;
}
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 drequires 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.
The first line contains a positive integer v (0 ≤ v ≤ 106). The second line contains nine positive integers a1, a2, ..., a9 (1 ≤ ai ≤ 105).
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.
5 5 4 3 2 1 2 3 4 5
55555
2 9 11 1 12 5 8 9 10 6
33
0 1 1 1 1 1 1 1 1 1
-1