permutation 2(递推 + 思维)

permutation 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)


Problem Description
You are given three positive integers N,x,y.
Please calculate how many permutations of 1N satisfies the following conditions (We denote the i-th number of a permutation by pi):

1. p1=x

2. pN=y

3. for all 1i<N|pipi+1|2
 

 

Input
The first line contains one integer T denoting the number of tests.

For each test, there is one line containing three integers N,x,y.

1T5000

2N105

1x<yN
 

 

Output
For each test, output one integer in a single line indicating the answer modulo 998244353.
 

 

Sample Input
3
4 1 4
4 2 4
100000 514 51144
 

 

Sample Output
2
1
253604680

 

算法:递推 + 思维

题解:你必须先考虑左端点和右端点,然后当左边和右边都填完了,之后便要x++,y--(是端点的话就不用),这是为什么呢就只有[x, y]个数了,然后填就有两种填法:x + 1 和  x + 2,之中x + 2就必须要返回填x + 1,再填x + 3,加上刚才的第一种可能 x + 1和第二种可能推出来的x + 3,这两个数又和最开始的x一样了,都又两种可能性,所以dp[i] = dp[i - 1] + dp[i - 3].

#include <iostream>
#include <cstdio>

using namespace std;

const int maxn = 1e5+5;
const int mod = 998244353;

int dp[maxn];

int main() {
    dp[0] = dp[1] = dp[2] = 1;
    for(int i = 3; i < maxn; i++) {
        dp[i] = (dp[i - 1] + dp[i - 3]) % mod;
    }
    int T;
    cin >> T;
    while(T--) {
        int n, x, y;
        scanf("%d %d %d", &n, &x, &y);
        if(x > 1) {     //不是端点的话就自加自减
            x++;
        }
        if(y < n) {
            y--;
        }
        int m = y - x;      //计算[x, y]区间中数的个数,不包括x, y
        printf("%d\n", dp[m]);
    }
    return 0;
}

 

posted @ 2019-08-05 16:47  不会fly的pig  阅读(310)  评论(0编辑  收藏  举报