URAL 1142——Relations——————【dp】

A - Relations

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice URAL 1142

Description

Background

Consider a specific set of comparable objects. Between two objects a and b, there exits one of the following three classified relations:

a = b
a < b
b < a

Because relation '=' is symmetric, it is not repeated above.

So, with 3 objects ( a, b, c), there can exist 13 classified relations:

a = b = c       a = b < c       c < a = b       a < b = c
b = c < a       a = c < b       b < a = c       a < b < c
a < c < b       b < a < c       b < c < a       c < a < b
c < b < a

Problem

Given N, determine the number of different classified relations between N objects.

Input

Includes many integers N (in the range from 2 to 10), each number on one line. Ends with −1.

Output

For each N of input, print the number of classified relations found, each number on one line.

Sample Input

input	output
2 3 -1	3 13

题目大意：给你n个人，让你给n个人排名，可以有并列，问你有多少种排名情况。

解题思路：定义dp[i][j]表示j个人有i个名次，那么dp[1][i] = 1，而dp[i][i] = (i!) i的阶乘。dp[i][j] = i*dp[i][j-1] + i*dp[i-1][j-1]。表示当第j个人加入的话，要么第j个人跟某些人等名次，要么有一个单独的名次。当第j个人跟其他人等名次的时候，他可以选择i种名次中的任意一种，当第j个人有单独的名次的时候，他可以在i-1种名次形成的i个空中选一个。   思路转自：http://www.zhihu.com/question/30200444?sort=created

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
typedef long long LL;
LL fact[20],dp[20][20];
void fac(){
    fact[0] = 1;
    for(LL i =1; i <= 15; i++){
        fact[i] = fact[i-1]*i;
    }
}
void prin(){
    for(int i = 1; i <= 12; i++){
        dp[1][i] = 1;
        dp[i][i] = fact[i];
    }
    for(int i = 2; i <= 12; i++){
        for(int j = i+1; j <= 12; j++){
            dp[i][j] = i*(dp[i][j-1] + dp[i-1][j-1]);
        }
    }
}
int main(){
    LL n;
    fac();
    prin();
    while(scanf("%lld",&n)!=EOF&&n!=-1){
        LL ans = 0;
        for(int i = 1; i <= n; i++){
            ans += dp[i][n];
        }
        printf("%lld
",ans);
    }
    return 0;
}

　　

uralID： 196348LD