首页 > 代码库 > XidianOJ 1183 Water Problem: Items divided
XidianOJ 1183 Water Problem: Items divided
题目描述
Youyouyouyou is very interested in math, one day, an idea came into his mind that how many ways he can patition n same things into no more than m groups? he think it too hard for him, so youyouyouyou ask wise cyt for help, but wise cyt don’t want to talk with youyouyouyou because it is too easy for him, now can you help youyouyouyou solve this problem?
输入
multi test cases, two integers n and m(1<=m<=n<=1000) , end with n = m = 0.
输出
output
#include <cstdio> #include <algorithm> #include <iostream> using namespace std; #define MOD 1000000007 typedef long long LL; LL f[1001][1001] = {0},ans[1001][1001] = {0}; int main(){ int n,m,i,j; for (i=1;i<=1000;i++){ f[i][0] = 0; f[i][1] = 1; f[i][i] = 1; f[0][i] = 0; } for (i=1;i<=1000;i++){ for (j=1;j<=i;j++){ if (i == j) f[i][j] = 1; else f[i][j] = (f[i-1][j-1] + f[i-j][j]) % MOD; } } for (i=1;i<=1000;i++){ ans[i][1] = f[i][1]; for (j=2;j<=i;j++){ ans[i][j] = (ans[i][j-1] + f[i][j]) % MOD; } } while (scanf("%d %d",&n,&m) != EOF){ if (n == 0 && m == 0) break; // LL res = 0; // for (i=1;i<=m;i++){ // res = (res + f[n][i]) % MOD; // } // printf("%lld\n",res); printf("%lld\n",ans[n][m]); } return 0; }
an answer modulo 1e9 + 7 per line
--正文
n个相同的球放入m个相同的盒子,允许有空盒
F(n,m) = F(n-1,m-1) + F(n-m,m)
注意边界的处理
XidianOJ 1183 Water Problem: Items divided
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。