1 1 2 3
1/2 6/5HintCase 1: S=‘01‘ or S=‘10‘, so the expected times = 1/2 = 1/2 Case 2: S=‘00011‘ or S=‘00101‘ or S=‘00110‘ or S=‘01001‘ or S=‘01010‘ or S=‘01100‘ or S=‘10001‘ or S=‘10010‘ or S=‘10100‘ or S=‘11000‘, so the expected times = (1+2+1+2+2+1+1+1+1+0)/10 = 12/10 = 6/5
用next_permutation将所有的全排列都打印出来，然后找结果，打表（反正结果也不大 12*12）；
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
const int M = 1e2+5;
const int INF = 0x3f3f3f3f;
using namespace std;
char str[15];
int c[12][12] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,
2, 6, 12, 20, 30, 42, 56, 72, 90, 110, 132, 156,
3, 12, 30, 60, 105, 168, 252, 360, 495, 660, 858, 1092,
4, 20, 60, 140, 280, 504, 840, 1320, 1980, 2860, 4004, 5460,
5, 30, 105, 280, 630, 1260, 2310, 3960, 6435, 10010, 15015, 21840,
6, 42, 168, 504, 1260, 2772, 5544, 10296, 18018, 30030, 48048, 74256,
7, 56, 252, 840, 2310, 5544, 12012, 24024, 45045, 80080, 136136, 222768,
8, 72, 360, 1320, 3960, 10296, 24024, 51480, 102960, 194480, 350064, 604656,
9, 90, 495, 1980, 6435, 18018, 45045, 102960, 218790, 437580, 831402, 1511640,
10, 110, 660, 2860, 10010, 30030, 80080, 194480, 437580, 923780, 1847560, 3527160,
11, 132, 858, 4004, 15015, 48048, 136136, 350064, 831402, 1847560, 3879876, 7759752,
12, 156, 1092, 5460, 21840, 74256, 222768, 604656, 1511640, 3527160, 7759752, 16224936};
int d[12][12] = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,
3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91,
4, 10, 20, 35, 56, 84, 120, 165, 220, 286, 364, 455,
5, 15, 35, 70, 126, 210, 330, 495, 715, 1001, 1365, 1820,
6, 21, 56, 126, 252, 462, 792, 1287, 2002, 3003, 4368, 6188,
7, 28, 84, 210, 462, 924, 1716, 3003, 5005, 8008, 12376, 18564,
8, 36, 120, 330, 792, 1716, 3432, 6435, 11440, 19448, 31824, 50388,
9, 45, 165, 495, 1287, 3003, 6435, 12870, 24310, 43758, 75582, 125970,
10, 55, 220, 715, 2002, 5005, 11440, 24310, 48620, 92378, 167960, 293930,
11, 66, 286, 1001, 3003, 8008, 19448, 43758, 92378, 184756, 352716, 646646,
12, 78, 364, 1365, 4368, 12376, 31824, 75582, 167960, 352716, 705432, 1352078,
13, 91, 455, 1820, 6188, 18564, 50388, 125970, 293930, 646646, 1352078, 2704156};
/*int f(char s[], int len){
int ans = 0;
for(int i = 1; i < len; ++ i)
if(s[i] == '1'&&s[i-1] == '0') ++ans;
return ans;
}
bool cmp(char a, char b){
return a < b;
}*/
int gcd(int a, int b){
if(a < b) swap(a, b);
if(b == 1) return 1;
int t = a%b;
while(t){
a = b; b = t;
t = a%b;
}
return b;
}
int main(){
int n, m;
while(cin >> n >>m){
int temp = gcd(d[n-1][m-1], c[n-1][m-1]);
printf("%d/%d\n", c[n-1][m-1]/temp, d[n-1][m-1]/temp);
}
/*freopen("D:\\hah.txt", "w", stdout); //打表；
for(int i = 1; i <= 12; ++ i){
for(int j = 1; j <= 12; ++ j){
int t = 0, s = i;
while(s --){
str[t++] = '0';
}
s = j;
while(s --){
str[t++] = '1';
}
str[t] = '\0';
c[i][j] = f(str, i+j); d[i][j] = 1;
while(next_permutation(&str[0], &str[i+j], cmp)){
//temp1++;
++d[i][j];
c[i][j] += f(str, i+j);
}
//d[i][j] = temp1;
//cout << i << " "<< j <<" " <<c[i][j] << "\n";
}
}
int t = 0;
for(int i = 1; i <= 12; ++ i){
for(int j = 1; j <= 12; ++ j){
cout << c[i][j] <<", ";
t++;
if(t %12 == 0) cout << "\n";
}
}
cout << endl <<endl <<endl;
t = 0;
for(int i = 1; i <= 12; ++ i){
for(int j = 1; j <= 12; ++ j){
cout << d[i][j] <<", ";
t++;
if(t %12 == 0) cout << "\n";
}
}
fclose(stdout);*/
return 0;
}
Hdoj 5194 DZY Loves Balls 【打表】+【STL】
原文地址：http://blog.csdn.net/shengweisong/article/details/44758695
