hduHolding Bin-Laden Captive!(母函数)

Holding Bin-Laden Captive!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18581    Accepted Submission(s): 8279

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

题目大意：有价值为1，2，5的三种硬币，其个数分别为num_1,num_2,num_5,问用这些硬币不能组合的面值中最小的是多少？

解题思路：开两个数组c1[]，c2[],先用母函数找出这些硬币可以表示的硬币的面值，随后循环找出第一个不能表示的面值。

AC代码：

#include<stdio.h>#include<string.h>int c1[1000010],c2[1000010];int main(){int a,b,c;int i,j,k,v;while(scanf("%d%d%d",&a,&b,&c)&&(a|b|c)){memset(c1,0,sizeof(c1));memset(c2,0,sizeof(c2));int d[3];d[0]=b;d[1]=c;int x[3]={2,5};int n=a+b*2+c*5;for(i=0;i<=a;i++)   //初始化母函数的第一表达式 c1[i]=1;for(i=0;i<2;i++){for(j=0;j<=2*n;j++){for(k=0;k*x[i]+j<=2*n&&k<=d[i];k++) c2[k*x[i]+j]+=c1[j];}for(j=0;j<=2*n;j++){c1[j]=c2[j];c2[j]=0;}}    v=0;    for(i=0;i<=2*n;i++)    //找出第一个方案数为零的就是其不能表示的最小面值；    {    if(c1[i]==0)    {    v=i;break;}}printf("%d\n",v);}return 0;}