timus 1017
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思路 :
给N个砖块。按要求求有多少种堆放的形式。
我们只考虑 N块砖 最后一列为J 块砖时 res[N][J] ( J = 1~ N - 1 ) 且易知res[N][N] = 1。
要求 res[N][J]等价于求 res[N - J][K] 有几种形式。( K = 1 ~ N - J && K < N - J && K < J )
因此方程为res[N][J] += res[N - J][K] ( K = 1 ~ N - J&& K < J )
#include <stdio.h> long long res[501][501];int n; int main ( ) { for ( int i = 1; i <= 500; ++i ) res[i][i] = 1; for ( int i = 2; i <= 500; ++i ) for ( int j = 1; j < i; ++j ) for ( int k = 1; k <= i - j; ++k ) if ( k < j ) res[i][j] += res[i - j][k]; while ( scanf ( "%d", &n ) != EOF ) { long long int sum = 0; for ( int i = 1; i < n; sum += res[n][i], ++i ) ; printf ( "%lld\n", sum ); } }
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