ICPC 2015 Changchun A Too Rich（贪心）

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.

输入

The ?rst 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 c i means how many coins/banknotes in denominations of i dollars in your wallet.
1≤T≤20000
0≤p≤109
0≤c i≤100000

输出

样例输入

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

样例输出

9 -1 36

题意：现在有 1，5,10,20,50,100,200,500,1000,2000面值的钱币，问你要凑够p元钱最多需要多少张纸币，给你p和每个面值纸币的数量
我的想法：但是WA了，首先，我会预处理出来到这个纸币最多能够构成的钱数，面值一共十种，那么我就二进制枚举这一种选还是不选，对于选的，我从后向前跑，对于这一种面值，我尽量少选，一保证前面可以
尽量多选，尽量少选就是最少选一张，最多选给定的数量张，在前面足够构成的情况下选。最后判断是否可以，但是WA了，我想应该是20,50,500，那里的问题，但是还没有找到反例。
找反例：20元的有4张，50元的有三张，这样的话去凑120元，按照我的想法，因为前面可以凑够80元，所以50的我只会选一张，剩下的70前面一定可以，但是事实上并不可以，OK，说服自己很舒服
于是正解：首先我们发现查安生错误的原因就是50和500需要多少张，为什么这两者特殊呢，因为对于别的数字，前面所有的数字都是这个数字的因子，唯独这两个数字前面的数字含有非因子，因此，对于每一个
数字，在我们找到最少选多少个之后，对于50,500，还要考虑要不要多选一张,其余的就不需要了。试着写一下
最原始想法代码：

#include<stdio.h> #include<iostream> #include<algorithm> #include<string.h> #include<math.h> using namespace std; int t; int num; int tmp; int a[13]; int c[13]; int sum[13]; int zhao[13] = {1,5,10,20,50,100,200,500,1000,2000}; int ans; int l; int fac[13]; void init() { fac[0] = 1; for(int i=1; i<=10; i++) fac[i] = fac[i-1]*2; } int main() { init(); scanf("%d", &t); while( t-- ) { scanf("%d", &num); ans = -1; for(int i=0; i<10; i++) { scanf("%d", &a[i]); if(i == 0) sum[i] = a[i]*zhao[i]; else sum[i] = sum[i-1]+a[i]*zhao[i]; if(sum[i] <= num) l = i+1; } for(int i=fac[l]-3; i<=1023; i++) { int ttmp = i; tmp = num; int res = 0; for(int j=9; j>=0; j--) { if(!(ttmp>>j)&1) continue; if(zhao[j] > tmp) { res = -1; break; } int wu = tmp-sum[j-1]; int liu = max(1, (int)ceil(1.0*wu/zhao[j])); liu = min(liu, a[j]); res += liu; tmp -= liu*zhao[j]; } if(tmp != 0) { res = -1; } if(res != -1) { ans = res; break; } } printf("%d\n", ans); } return 0; }

#include<stdio.h> #include<iostream> #include<algorithm> #include<string.h> #include<math.h> using namespace std; const int maxn = 13; int t; int num; int a[maxn]; int f[10] = {1,5,10,20,50,100,200,500,1000,2000}; int sum[13]; int ans; void init() { ans = -1; } void input() { scanf("%d" , &num); for(int i=0; i<10; i++) { scanf("%d" , &a[i]); if(i == 0) sum[i] = a[i]*f[i]; else sum[i] = sum[i-1]+a[i]*f[i]; // printf("%d..%d..\n" , i , sum[i]); } } void solve(int id, int rem , int res) { // printf("%d..%d..%d..\n" , id , rem , res); if(id < 0) { if(rem == 0) { ans = max(ans , res); } return ; } int tmp = rem-sum[id-1]; int ttmp = max(0,(int)ceil(1.00*tmp/f[id])); ttmp = min(ttmp , a[id]); ///算出来选择多少张 solve(id-1 , rem-f[id]*ttmp , res+ttmp); if(f[id]==50 || f[id]==500) { ttmp++; if(ttmp <= a[id]) solve(id-1 , rem-f[id]*ttmp , res+ttmp); } } int main() { scanf("%d" , &t); while( t-- ) { init(); input(); solve(9,num,0); printf("%d\n" , ans); } return 0; }

看题解上写的是转换成尽可能多的去掉大面值的，使得剩下的可以构成要求的数字，对于50与500，另外考虑下少去掉一张，与我写的尽可能少选择大面值的，多选择一张有什么不同之处吗

#include<stdio.h> #include<iostream> #include<algorithm> #include<string.h> #include<math.h> using namespace std; const int maxn = 13; const int inf = 10000000; int t; int num; int a[maxn]; int f[10] = {1,5,10,20,50,100,200,500,1000,2000}; int sum[13]; int ans; int all; void init() { all = 0; ans = 10000000; } void input() { scanf("%d" , &num); for(int i=0; i<10; i++) { scanf("%d" , &a[i]); all += a[i]; if(i == 0) sum[i] = a[i]*f[i]; else sum[i] = sum[i-1]+a[i]*f[i]; } } void solve(int id, int rem , int res) { if(id < 0) { if(rem == 0) { ans = min(ans , res); } return ; } int tmp = a[id]; tmp = min(tmp , rem/f[id]); tmp = max(0 , tmp); solve(id-1 , rem-tmp*f[id] , res+tmp); if(tmp) { tmp--; solve(id-1 , rem-tmp*f[id] , res+tmp); } } int main() { scanf("%d" , &t); while( t-- ) { init(); input(); solve(9,sum[9]-num,0); // printf("%d...\n" , ans); if(ans == 10000000) printf("-1\n"); else printf("%d\n" , all-ans); } return 0; }

ICPC 2015 Changchun A Too Rich（贪心）

问题 A: Too Rich

题目描述

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输出

样例输入

样例输出