动态规划,状态转移方程dp[i][j] = dp[i - 1][j] + dp[i - 2][j - 1]其中i表示总数,j表示取出的数个数。初始化时要注意顺序问题,设取出个数为k,则所有2k-1的项都应当为0,但特殊情况是取出一个时有几个数就输出几个。另外一个特殊情况就是0,直接输出空集1。
// Problem#: 4428
// Submission#: 3295621
// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License
// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/
// All Copyright reserved by Informatic Lab of Sun Yat-sen University
#include<iostream>
#include<cstdio>
using namespace std;
int dp[1000][1000];
int main() {
int n, m, test;
scanf("%d", &test);
while (test--) {
scanf("%d%d", &n, &m);
if (m == 0) {
printf("1\n");
continue;
}
for (int i = 0; i < 2 * m - 1; i++) {
for (int j = 0; j <= m; j++) {
dp[i][j] = 0;
}
}
for (int i = 1; i <= n; i++) {
dp[i][1] = i;
}
for (int i = 2; i <= n; i++) {
for (int j = 2; j <= m; j++) {
dp[i][j] = dp[i - 1][j] + dp[i - 2][j - 1];
}
}
printf("%d\n", dp[n][m]);
}
}