ICPC 2015 Changchun A Too Rich(貪心)
阿新 • • 發佈:2018-10-04
其余 技術分享 ice 需要 icpc 原因 要求 while 預處理
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.
1≤T≤20000
0≤p≤109
0≤c i≤100000
問題 A: Too Rich
時間限制: 1 Sec 內存限制: 128 MB題目描述
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.
輸入
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
輸出
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‘.
樣例輸入
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
樣例輸出
9 -1 36
題意:現在有 1,5,10,20,50,100,200,500,1000,2000面值的錢幣,問你要湊夠p元錢最多需要多少張紙幣,給你p和每個面值紙幣的數量
我的想法:但是WA了,首先,我會預處理出來到這個紙幣最多能夠構成的錢數,面值一共十種,那麽我就二進制枚舉這一種選還是不選,對於選的,我從後向前跑,對於這一種面值,我盡量少選,一保證前面可以
盡量多選,盡量少選就是最少選一張,最多選給定的數量張,在前面足夠構成的情況下選。最後判斷是否可以,但是WA了,我想應該是20,50,500,那裏的問題,但是還沒有找到反例。
找反例:20元的有4張,50元的有三張,這樣的話去湊120元,按照我的想法,因為前面可以湊夠80元,所以50的我只會選一張,剩下的70前面一定可以,但是事實上並不可以,OK,說服自己很舒服
於是正解:首先我們發現查安生錯誤的原因就是50和500需要多少張,為什麽這兩者特殊呢,因為對於別的數字,前面所有的數字都是這個數字的因子,唯獨這兩個數字前面的數字含有非因子,因此,對於每一個
數字,在我們找到最少選多少個之後,對於50,500,還要考慮要不要多選一張,其余的就不需要了。試著寫一下
最原始想法代碼:
#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; }View Code
錯一發代碼:
#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]); ///算出來選擇多少張 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; }View Code
看題解上寫的是轉換成盡可能多的去掉大面值的,使得剩下的可以構成要求的數字,對於50與500,另外考慮下少去掉一張,與我寫的盡可能少選擇大面值的,多選擇一張有什麽不同之處嗎
代碼是沒有什麽問題的 應該是邏輯上的問題
先改成去掉寫一下試試
為什麽就過了呢 這個問題暫時挖坑
#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(貪心)