Holding Bin-Laden Captive!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 14670 Accepted Submission(s): 6566
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
Author
lcy
母函数的变形。求出来每一个的种类数，存入数组中，之后再推断是否为零即可了，
当然也有个简单的办法，就是观察是否可以超出四。这样更简单，代码更短。可是。就是不太easy想得出来：
简单的方法的代码例如以下：
#include int main(){ int a,b,c; while(~scanf("%d%d%d",&a,&b,&c),a+b+c) { if(a==0) { printf("1\n"); } else if(a+2*b<4) { printf("%d\n",a+b*2+1); } else printf("%d\n",a+2*b+5*c+1); } return 0;} 至于母函数的代码也贴出来例如以下：
#include int c1[10100],c2[10100];int main(){ int i,j,a,b,c; while(~scanf("%d%d%d",&a,&b,&c),a+b+c) { int sum=a+b*2+c*5; for(i=0;i<=sum;i++) c1[i]=c2[i]=0; for(i=0;i<=a;i++) c1[i]=1; for(i=0;i<=a;i++) for(j=0;j<=2*b;j+=2) c2[i+j]+=c1[i]; for(j=0;j<=sum;j++) c1[j]=c2[j],c2[j]=0; for(i=0;i<=a+2*b;i++) for(j=0;j<=5*c;j+=5) c2[i+j]+=c1[i];//起初在这个地方。将i写成了j结果wa了好几回 for(j=0;j<=a+2*b+5*c;j++) c1[j]=c2[j],c2[j]=0; for(j=0;j<=sum;j++) { if(c1[j]==0) { printf("%d\n",j); break; } } if(j>=sum+1) printf("%d\n",j); } return 0;}