HDU 1085 Holding Bin-Laden Captive!

Holding Bin-Laden Captive!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16585 Accepted Submission(s): 7451


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 3
0 0 0


Sample Output

4

给出1 2 5硬币各有多少个，问不能凑成的钱的最小值是多少。a[i]里存的是当钱数为i时，拿现有的硬币能够凑成i元的方法，凑成i元，一共有a[i]中方法。

#include <iostream>#include <cstdio>#include <memory.h>using namespace std;int a[10000],b[10000];void mother(int i,int num,int n){//i代表第i个元素，第i个元素的数量有num个，需要计算的最大指数n    for(int j=0;j<=n;j++)        for(int k=0;k<=num&&k*i+j<=n;k++)        b[k*i+j]+=a[j];    for(int j=0;j<=n;j++)    {        a[j]=b[j];        b[j]=0;    }}int main(void){    int c2[5];    int c1[5];    c1[1]=1;    c1[2]=2;    c1[3]=5;    while(scanf("%d%d%d",&c2[1],&c2[2],&c2[3]),(c2[1]||c2[2]||c2[3]))    {        int sum=c2[1]+2*c2[2]+5*c2[3]+1;        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        a[0]=1;        for(int i=1;i<=3;i++)            mother(c1[i],c2[i],sum);           for(int i=1;i<=sum;i++)           {               if(a[i]==0)               {                    printf("%d\n",i);                    break;               }           }    }    return 0;}