HDU 1085 Holding Bin-Laden Captive! 母函数

点击打开链接

Holding Bin-Laden Captive!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13183    Accepted Submission(s): 5907

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

//484MS312K#include<stdio.h>#include<string.h>#define M 10007int c1[M],c2[M],num[4],id[4];int main(){    int a,b,c;    while(scanf("%d%d%d",&a,&b,&c),a|b|c)    {        memset(num,0,sizeof(num));        num[1]=a;num[2]=b;num[3]=c;        memset(c1,0,sizeof(c1));//c1保存当前得到的多项式各项系数        memset(c2,0,sizeof(c2));//c2保存每次计算时临时的结果        c1[0]=1;  //x^0去乘后面多项式        id[1]=1;id[2]=2;id[3]=5;        for(int i=1;i<=3;i++)//要乘以3个多项式        {            for(int j=0;j<=M;j++)//c1的各项指数                for(int k=0;k<=num[i]&&j+k*id[i]<=M;k++)//k*i表示被乘多项式各项的指数,(X^(0*i) + X^(1*i) + X^(2*i) + ……)                    c2[j+k*id[i]]+=c1[j];//指数相加得j+k*i,加多少只取决于c1[j]的系数，因为被乘多项式的各项系数均为1            memcpy(c1,c2,sizeof(c2));            memset(c2,0,sizeof(c2));        }       for(int i=1;i<=M;i++)        if(c1[i]==0)        {            printf("%d\n",i);break;        }    }    return 0;}