2019独角兽企业重金招聘Python工程师标准>>> 
Holding Bin-Laden Captive!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 15010 Accepted Submission(s): 6726
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!
“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
题意：题目给出一些钱的数量，要求用给出的钱组合不能组成的最小钱数；
意解：母函数，把组合加法转化成幂的加法；
AC代码：
题意：题目给出一些钱的数量，要求用给出的钱组合不能组成的最小钱数；
意解：母函数，把组合加法转化成幂的加法；
AC代码：
- #include <cstdio>
- #include <cstring>
- #define M 8005
- int vis[M];//标记可以组合的数；
- int main()
- {
- int num_1,num_2,num_5;
- while(true)
- {
- scanf("%d %d %d",&num_1,&num_2,&num_5);
- if(num_1 + num_2 + num_5 == 0) break;
- memset(vis,0,sizeof(vis));
- for(int i = 0; i <= num_1; i++)
- vis[i] = 1;
- for(int j = 0; j <= num_2 * 2; j += 2)
- {
- for(int k = 0; k <= num_1; k++)
- {
- vis[j + k] = 1;
- }
- }
- for(int j = 0; j <= num_5 * 5; j += 5)
- {
- for(int k = 0; k <= num_1 + num_2 * 2; k++)
- {
- if(vis[k]) //如果当前数可以组合，则加上；
- {
- vis[k + j] = 1;
- }
- }
- }
- int i;
- for(i = 1; i <= num_1 + num_2 * 2 + num_5 * 5; i++)
- {
- if(vis[i] == 0)
- {
- break;
- }
- }
- printf("%d\n",i);
- }
- return 0;
- }