Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7479 Accepted Submission(s):
5313
Problem Description
"Well, it seems the first problem is too easy. I will
let you know how foolish you are later." feng5166 says.
"The second problem is, given an positive integer N, we define an equation like this:
N=a[1]+a[2]+a[3]+...+a[m];
a[i]>0,1<=m<=N;
My question is how many different equations you can find for a given N.
For example, assume N is 4, we can find:
4 = 4;
4 = 3 + 1;
4 = 2 + 2;
4 = 2 + 1 + 1;
4 = 1 + 1 + 1 + 1;
so the result is 5 when N is 4. Note that "4 = 3 + 1" and "4 = 1 + 3" is the same in this problem. Now, you do it!"
"The second problem is, given an positive integer N, we define an equation like this:
N=a[1]+a[2]+a[3]+...+a[m];
a[i]>0,1<=m<=N;
My question is how many different equations you can find for a given N.
For example, assume N is 4, we can find:
4 = 4;
4 = 3 + 1;
4 = 2 + 2;
4 = 2 + 1 + 1;
4 = 1 + 1 + 1 + 1;
so the result is 5 when N is 4. Note that "4 = 3 + 1" and "4 = 1 + 3" is the same in this problem. Now, you do it!"
Input
The input contains several test cases. Each test case
contains a positive integer N(1<=N<=120) which is mentioned above. The
input is terminated by the end of file.
Output
For each test case, you have to output a line contains
an integer P which indicate the different equations you have found.
Sample Input
4
10
20
Sample Output
5
42
627
1 /* 2 copy: 3 首先,我们引进一个小小概念来方便描述吧,record[n][m]是把自然数n划分成所有元素不大于m的分法,例如: 4 当n=4,m=1时,要求所有的元素都比m小,所以划分法只有1种:{1,1,1,1}; 5 当n=4,m=2时,。。。。。。。。。。。。。。。。只有3种{1,1,1,1},{2,1,1},{2,2}; 6 当n=4,m=3时,。。。。。。。。。。。。。。。。只有4种{1,1,1,1},{2,1,1},{2,2},{3,1}; 7 当n=4,m=5时,。。。。。。。。。。。。。。。。只有5种{1,1,1,1},{2,1,1},{2,2},{3,1},{4}; 8 从上面我们可以发现:当n==1||m==1时,只有一种分法; 9 当n<m时,由于分法不可能出现负数,所以record[n][m]=record[n][n]; 10 当n==m时,那么就得分析是否要分出m这一个数,如果要分那就只有一种{m},要是不分,那就是把n分成不大于m-1的若干份; 11 即record[n][n]=1+record[n][n-1]; 12 当n>m时,那么就得分析是否要分出m这一个数,如果要分那就{{m},{x1,x2,x3..}}时n-m的分法record[n-m][m], 13 要是不分,那就是把n分成不大于m-1的若干份;即record[n][n]=record[n-m][m]+record[n][m-1]; 14 那么其递归式: 15 record[n][m]= 1 (n==1||m==1) 16 record[n][n] (n<m) 17 1+record[n][m-1] (n==m) 18 record[n-m][m]+record[n][m-1] (N>m) 19 */ 20 #include<iostream> 21 using namespace std; 22 23 int main() 24 { 25 int i,j,n,a[125][125]={0}; 26 for(i=1;i<125;i++) 27 a[i][1]=a[1][i]=1; 28 for(i=2;i<125;i++) 29 for(j=2;j<125;j++) 30 { 31 if(i<j) 32 a[i][j]=a[i][i]; 33 if(i==j) 34 a[i][j]=a[i][j-1]+1; 35 if(i>j) 36 a[i][j]=a[i-j][j]+a[i][j-1]; 37 } 38 while(scanf("%d",&n)!=EOF) 39 printf("%d\n",a[n][n]); 40 return 0; 41 }
功不成,身已退