HDU 5527 Too Rich

Too Rich

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 395 Accepted Submission(s): 118

Problem Description
You are a rich person, and you think your wallet is too heavy and full now. So you want to give me some money by buying a lovely pusheen sticker which costs p dollars from me. To make your wallet lighter, you decide to pay exactly p dollars by as many coins and/or banknotes as possible.

For example, if p=17 and you have two $10 coins, four $5 coins, and eight $1 coins, you will pay it by two $5 coins and seven $1 coins. But this task is incredibly hard since you are too rich and the sticker is too expensive and pusheen is too lovely, please write a program to calculate the best solution.

Input
The first line contains an integer T indicating the total number of test cases. Each test case is a line with 11 integers p,c1,c5,c10,c20,c50,c100,c200,c500,c1000,c2000, specifying the price of the pusheen sticker, and the number of coins and banknotes in each denomination. The number ci means how many coins/banknotes in denominations of i dollars in your wallet.

1≤T≤20000
0≤p≤109
0≤ci≤100000

Output
For each test case, please output the maximum number of coins and/or banknotes he can pay for exactly p dollars in a line. If you cannot pay for exactly p dollars, please simply output '-1'.

[align=left]Sample Input[/align]

3
17 8 4 2 0 0 0 0 0 0 0
100 99 0 0 0 0 0 0 0 0 0
2015 9 8 7 6 5 4 3 2 1 0

[align=left]Sample Output[/align]

9

-1

36

[align=left]Source[/align]
2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）

解题：
看看hqwhqwhq的解法
贪心，用尽量多的小面值的纸币来凑数，基本上每一个数都能整除后面一个数，所以前i种面值最多能凑出$sum[i]$元，那么第$i+1$种面值需要$\frac{p-sum[i]}{cnt[i+1]}$
此时有个小问题就是，20并不能整除50，但是20|2∗50，所以当面值为50,500的时候，可能需要多取一张。

#include <bits/stdc++.h>
using namespace std;
using LL = long long;
const int maxn = 11;
const int val[] = {0, 1, 5, 10, 20, 50, 100, 200, 500, 1000, 2000};
int cnt[maxn],ret;
LL sum[maxn];
void dfs(LL rest,int pos,int cnt){
if(rest < 0) return;
if(!pos){
if(!rest) ret = max(ret,cnt);
return;
}
LL tmp = max(0LL,rest - sum[pos-1]);
int tnt = tmp/val[pos];
if(tmp%val[pos]) ++tnt;
if(tnt <= ::cnt[pos]) dfs(rest - (LL)tnt*val[pos],pos - 1,cnt + tnt);
if(++tnt <= ::cnt[pos]) dfs(rest - (LL)tnt*val[pos],pos - 1,cnt + tnt);
}
int main(){
int kase,money;
scanf("%d",&kase);
while(kase--){
scanf("%d",&money);
for(int i = 1; i < maxn; ++i){
scanf("%d",cnt + i);
sum[i] = sum[i - 1] + static_cast<LL>(cnt[i])*val[i];
}
ret = -1;
dfs(money,10,0);
printf("%d\n",ret);
}
return 0;
}

View Code