ACM-ICPC 2017 Asia Xi'an J LOL 【暴力 && 排列组合】

任意门：https://nanti.jisuanke.com/t/20750

J - LOL

5 friends play LOL together . Every one should BAN one character and PICK one character . The enemy should BAN 55 characters and PICK 55 characters . All these 2020 heroes must be different .

Every one can BAN any heroes by his personal washes . But he can only PICK heroes which he has bought .

Suppose the enemy can PICK or BAN any heroes. How many different ways are there satisfying the conditions?

For example , a valid way is :

Player 11 : picks hero 11, bans hero 22

Player 22 : picks hero 33, bans hero 44

Player 33 : picks hero 5, bans hero 66

Player 44 : picks hero 77, bans hero 88

Player 55 : picks hero 99, bans hero 1010

Enemies pick heroes 11,12,13,14,1511,12,13,14,15 , ban heroes 16,17,18,19,2016,17,18,19,20 .

Input

The input contains multiple test cases.(No more than 2020)

In each test case . there’s 55 strings S[1] \sim S[5]S[1]∼S[5] ,respectively whose lengths are 100100 , For the ii-th person if he has bought the jj-th hero, the jj-th character of S[i]S[i] is '11', or '00' if not. The total number of heroes is exactly 100100 .

Output

For each test case , print the answer mod 10000000071000000007 in a single line .

样例输入复制

0110011100011001001100011110001110001110001010010111111110101010010011010000110100011001001111101011

1000111101111110110100001101001101010001111001001011110001111110101000011101000001011100001001011010

0100101100011110011100110110011100111100010010011001111110101111111000000110001110000110001100001110

1110010101010001000110100011101010001010000110001111111110101010000000001111001110110101110000010011

1000010011111110001101100000101001110100011000111010011111110110111010011111010110101111011111011011

样例输出复制

515649254

题意概括：

五个二进制数代表我方 5 人所拥有的英雄，0为没有，1为有。

敌方5人什么英雄都有，问有多少种 Ban Pick方案。

每人都可以选一个自己拥有的英雄和 ban 掉任何一个英雄，最后选的和禁止掉的20个英雄都不相同。

解题思路：

暴力枚举我方选英雄的方案

敌方选英雄的方案为 A（95， 5）

我方禁英雄的方案为 C（90， 5）

敌方禁英雄的方案为 C（85，5）

关于暴力的优化和剪枝：

如果是传统暴力枚举每个人选不同英雄的方案，需要五层 for（0， 100）循环，TL！

优化只枚举前四个人的方案，第五个能选的英雄只能是前面没有选的，缩小了一层循环。

关于剪枝前面选过的不要选咯

AC code：

 #include <cstdio>

 #include <iostream>

 #include <algorithm>

 #include <cstring>

 #define INF 0x3f3f3f3f

 #define LL long long

 using namespace std;

 const int mod = 1e9+;

 const int MAXN = ;

 char a[MAXN], b[MAXN], c[MAXN], d[MAXN], e[MAXN];

 LL A(LL N, LL R)

 {

     LL sum = ;

     for(int i = ; i < R; i++)

         sum *= N-i;

     return sum;

 }

 LL C(LL N, LL R)

 {

     LL sum = ;

     for(int i = ; i <= R; i++)

         sum = sum* (N+-i)/i;

     return sum;

 }

 int main()

 {

     while(~scanf("%s%s%s%s%s", a, b, c, d, e)){

         int len = ;

         for(int i = ; i < ; i++){

             if(e[i] == '') len++;

         }

         LL ans = ;

         for(int i = ; i < ; i++){

             if(a[i] == '') continue;

             for(int j = ; j < ; j++){

                 if(b[j] == '' || j == i) continue;

                 for(int k = ; k < ; k++){

                     if(c[k] == '' || k == i || k == j) continue;

                     for(int p = ; p < ; p++){

                         if(d[p] == '' || p == i || p == j || p == k) continue;

                         int res = len;

                         if(e[i] == '') res--;

                         if(e[j] == '') res--;

                         if(e[k] == '') res--;

                         if(e[p] == '') res--;

                         if(res>=)      ans+=res;

                     }

                 }

             }

         }

         ans = ans*A(, )%mod*C(, )%mod*C(, )%mod;

         printf("%lld\n", ans);

     }

     return ;

 }