hdoj1028 Ignatius and the Princess III（整数分解）

Ignatius and the Princess III

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)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!"InputThe 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.OutputFor each test case, you have to output a line contains an integer P which indicate the different equations you have found.Sample Input

41020

Sample Output

542627并未找出递推式，然后做法有点傻，数据太大，单纯递归肯定TLE；先用递归打印出前120项（需要运行大约6分钟orz，递归效率低，等的心好累），复制下来再直接打表。。。。。。。。。。代码：
#include<stdio.h>int k,z;void f(int n,int m){if(n==0||m==1){k++;z=0;return ;}int i;for(i=z;i<=n/m;i++){z=i;f(n-i,m-1);}}int main(){int n;int s[130]={0,1,2,3,5,7,11,15,22,30,42,56,77,101,135,176,231,297,385,490,627,792,1002,1255,1575,1958,2436,3010,3718,4565,5604,6842,8349,10143,12310,14883,17977,21637,26015,31185,37338,44583,53174,63261,75175,89134,105558,124754,147273,173525,204226,239943,281589,329931,386155,451276,526823,614154,715220,831820,966467,1121505,1300156,1505499,1741630,2012558,2323520,2679689,3087735,3554345,4087968,4697205,5392783,6185689,7089500,8118264,9289091,10619863,12132164,13848650,15796476,18004327,20506255,23338469,26543660,30167357,34262962,38887673,44108109,49995925,56634173,64112359,72533807,82010177,92669720,104651419,118114304,133230930,150198136,169229875,190569292,214481126,241265379,271248950,304801365,342325709,384276336,431149389,483502844,541946240,607163746,679903203,761002156,851376628,952050665,1064144451,1188908248,1327710076,1482074143,1653668665,1844349560};int i;for(i=1;i<=10;i++){k=0;f(i,i);}while(scanf("%d",&n)!=EOF){k=0;z=0;printf("%d\n",s);}return 0;}