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 ;
}
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