UVa 10943 - How do you add?
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题目:求一个数字n拆成k个数字的拆法数,可以重复,可以有0。
分析:dp,组合数学。
方法1:dp
状态:f(i,j)为 j 拆成 i 个数字的方法数,则有f(i,j)= sum(f(i,k)) { 0 ≤ k ≤ j };
方法2:计数原理
隔板法:C(n+k-1,k-1)= (n+1)(n+2)... (n+k-1),
计算利用 C(n,m)= C(n-1,m-1)+ C(n-1,m)即可。
说明:╮(╯▽╰)╭。
方法一:dp
#include <iostream>#include <cstdlib>#include <cstdio>using namespace std;int F[101][101];int main(){for ( int i = 0 ; i < 101 ; ++ i )for ( int j = 0 ; j < 101 ; ++ j )F[i][j] = 0;for ( int i = 0 ; i < 101 ; ++ i )F[1][i] = 1;for ( int i = 1 ; i < 101 ; ++ i )for ( int j = 0 ; j < 101 ; ++ j )for ( int k = 0 ; k <= j ; ++ k )F[i][j] = (F[i][j]+F[i-1][j-k])%1000000;int n,m;while ( scanf("%d%d",&n,&m) && n+m )printf("%d\n",F[m][n]);return 0;}方法二:计数原理
#include <iostream>#include <cstdlib>#include <cstdio>using namespace std;int C[201][201];int main(){for ( int i = 0 ; i < 201 ; ++ i )for ( int j = 0 ; j < 201 ; ++ j )C[i][j] = 0;for ( int i = 0 ; i < 201 ; ++ i )C[i][0] = 1;for ( int i = 1 ; i < 201 ; ++ i )for ( int j = 1 ; j <= i ; ++ j )C[i][j] = (C[i-1][j]+C[i-1][j-1])%1000000;int n,m;while ( scanf("%d%d",&n,&m) && n+m )printf("%d\n",C[n+m-1][m-1]);return 0;}
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