C++博客-Yiner的ACM-随笔分类-母函数

母函数~Fruit

Yiner — Tue, 15 Feb 2011 10:33:00 GMT

Fruit

Time Limit : 1000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 78 Accepted Submission(s) : 47

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Problem Description

转眼到了收获的季节，由于有TT的专业指导，Lele获得了大丰收。特别是水果，Lele一共种了N种水果，有苹果，梨子，香蕉，西瓜……不但味道好吃，样子更是好看。

于是，很多人们慕名而来，找Lele买水果。

甚至连大名鼎鼎的HDU ACM总教头 lcy 也来了。lcy抛出一打百元大钞，"我要买由M个水果组成的水果拼盘，不过我有个小小的要求，对于每种水果，个数上我有限制，既不能少于某个特定值，也不能大于某个特定值。而且我不要两份一样的拼盘。你随意搭配，你能组出多少种不同的方案，我就买多少份！"

现在就请你帮帮Lele，帮他算一算到底能够卖出多少份水果拼盘给lcy了。

注意，水果是以个为基本单位，不能够再分。对于两种方案，如果各种水果的数目都相同，则认为这两种方案是相同的。

最终Lele拿了这笔钱，又可以继续他的学业了～

Input

本题目包含多组测试，请处理到文件结束(EOF)。
每组测试第一行包括两个正整数N和M(含义见题目描述，0接下来有N行水果的信息，每行两个整数A,B(0<=A<=B<=100)，表示至少要买该水果A个，至多只能买该水果B个。

Output

对于每组测试，在一行里输出总共能够卖的方案数。
题目数据保证这个答案小于10^9

Sample Input

Sample Output

2
12

Author

Linle

Source

ACM程序设计期末考试——2008-01-02（3 教417）

1#include<stdio.h>
2int main()
3{
4    int i,n,m,j,k,a[101],b[101],c1[101],c2[101];
5    while(scanf("%d %d",&n,&m)!=EOF)
6    {
7        for(i=0;i<n;i++)
8        scanf("%d %d",&a[i],&b[i]);
9        for(i=0;i<=m;i++)
10        {
11            c1[i]=0;
12            c2[i]=0;
13        }
14        c1[0]=1;//第一次结束以后 k的就被赋值我为1了可其余还为0 就算c1[j]中被用到还是没效果只有1的c1[j]才会有效果
15        for(i=0;i<n;i++)
16        {
17            for(j=0;j<=m;j++)
18         for(k=a[i];k+j<=m&&k<=b[i];k++)
19               c2[j+k]+=c1[j];
20           for(j=0;j<=m;j++)
21        {
22            c1[j]=c2[j];
23            c2[j]=0;
24        }
25        }
26        printf("%d\n",c1[m]);
27    }
28    return 0;
29}

Yiner 2011-02-15 18:33 发表评论

母函数~Holding Bin-Laden Captive!

Yiner — Tue, 15 Feb 2011 10:31:00 GMT

Holding Bin-Laden Captive!

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 132 Accepted Submission(s) : 69

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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

Author

lcy

#include<stdio.h>
#include<string.h>
int main()
{
    int i,j,k,a,b,c,d[10001];
    while(scanf("%d %d %d",&a,&b,&c)!=EOF)
    {
        if(0==a&&0==b&&0==c)
        break;
        memset(d,0,sizeof(d));
          for(j=0;j<=a;j++)
          for(k=0;k+j<=a+2*b;k=k+2)
             d[k+j]++;
             for(j=0;j<=a+2*b&&d[j];j++)//&&a[j]是排除从0到a+2b之内不存在的项
             for(k=0;k+j<=a+2*b+5*c;k=k+5)
              d[k+j]++;
              for(i=0;i<=a+2*b+5*c+1;i++)//+1是因为如果情况中的全满足那么大于范围的第一个就是所求
             {
                 if(d[i]==0)
                 {
                     printf("%d\n",i);
                     break;
                 }
             }
    }
    return 0;
}

Yiner 2011-02-15 18:31 发表评论

母函数~Square Coins

Yiner — Tue, 15 Feb 2011 10:29:00 GMT

摘要: 阅读全文

Yiner 2011-02-15 18:29 发表评论

母函数~Ignatius and the Princess III

Yiner — Tue, 15 Feb 2011 10:28:00 GMT

摘要: 阅读全文

Yiner 2011-02-15 18:28 发表评论