hdu 1085

Holding Bin-Laden Captive!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15285    Accepted Submission(s): 6848

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

 

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问题分析：这一题老师超时，结果将 k*coin[i-1]+j<=sum&&k<=num[i];k++)该过之后就好了
这里注意每一个括号的倍数不一样时，这时就要注意用另一个数组来存储每个括号的项数
这样就可以灵活的运用母函数了 
#include<stdio.h>
#include<string.h>
int c1[8010]={0},c2[8010]={0};//用来存储括号多项式里的系数 
int main()
{
int num[3]={0},i,j,k,sum;//sum是用来确保要的最大指数，用来控制操作的复杂度，一定要注意 
int coin[3]={1,2,5};//存储相应括号的值 
while(scanf("%d%d%d",&num[1],&num[2],&num[3]),num[1]+num[2]+num[3])
{
sum=num[1]+num[2]*2+num[3]*5+2;
memset(c2,0,sizeof(c2));
memset(c1,0,sizeof(c1));
c1[0]=1;
if(num[1]==0) 
{
printf("1\n");
continue;
}
for(i=1;i<=3;i++)
{
for(j=0;j<=sum;j++)//sum的控制作用 
{
for(k=0;k*coin[i-1]+j<=sum&&k<=num[i];k++)//k*coin[i-1]+j<=sum&&很重要 
{
c2[k*coin[i-1]+j]+=c1[j];
}
}
for(j=0;j<=sum;j++)
{
c1[j]=c2[j];
c2[j]=0;
}
}
for(i=0;i<sum&&c1[i]!=0;i++);
printf("%d\n",i);
memset(num,0,sizeof(num));
}
return 0;
}