首页 > 代码库 > Holding Bin-Laden Captive!（1.多重背包 2.母函数）
Holding Bin-Laden Captive!（1.多重背包 2.母函数）
Holding Bin-Laden Captive! |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 41 Accepted Submission(s): 29 |
Problem Description We all know that Bin-Laden is a notorious terrorist, and he has disappeared for a long time. But recently, it is reported that he hides in Hang Zhou of China! “Oh, God! How terrible! ”  Don’t be so afraid, guys. Although he hides in a cave of Hang Zhou, he dares not to go out. Laden is so bored recent years that he fling himself into some math problems, and he said that if anyone can solve his problem, he will give himself up! Ha-ha! Obviously, Laden is too proud of his intelligence! But, what is his problem? “Given some Chinese Coins (硬币) (three kinds-- 1, 2, 5), and their number is num_1, num_2 and num_5 respectively, please output the minimum value that you cannot pay with given coins.” You, super ACMer, should solve the problem easily, and don’t forget to take $25000000 from Bush! |
Input Input contains multiple test cases. Each test case contains 3 positive integers num_1, num_2 and num_5 (0<=num_i<=1000). A test case containing 0 0 0 terminates the input and this test case is not to be processed. |
Output Output the minimum positive value that one cannot pay with given coins, one line for one case. |
Sample Input 1 1 30 0 0 |
Sample Output 4 |
Author lcy |
/*题意：给你面值1，2，5的金币的个数，让你求出最小不能组成的金额初步思路：暴力是不可取 1e12，以前做过一道丑数的题，这个题可以拿来参考一下#错误：每个能凑成的数都是由前一个状态转移过来的 根本行不通，那个是没有数量限制的......多重背包再来搞一下，总共三种物品每个取或者不取，没有覆盖的点就是不能凑出来的 #感悟：......我去，竟然多重背包爆过了*/#include<bits/stdc++.h>using namespace std;int num[4];int dp[8010];//dp[i]表示i是否能凑出来int cur=0;int v[]={1,2,5};void init(){ memset(dp,0,sizeof dp);}int main(){ //freopen("in.txt","r",stdin); while(scanf("%d%d%d",&num[0],&num[1],&num[2])!=EOF&&(num[0]+num[1]+num[2])){ init(); cur=num[0]+num[1]*2+num[2]*5;//总的钱数 dp[0]=1; for(int i=0;i<3;i++){ for(int j=0;j<num[i];j++){ for(int k=cur;k>=v[i];k--){ dp[k]+=dp[k-v[i]]; } } } int id=cur+1; for(int i=0;i<cur;i++){ //cout<<dp[i]<<" "; if(!dp[i]){ id=i; break; } } printf("%d\n",id); //cout<<endl; } return 0;}
未完......待续
Holding Bin-Laden Captive!（1.多重背包 2.母函数）
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。