problem description
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
0110011100011001001100011110001110001110001010010111111110101010010011010000110100011001001111101011 1000111101111110110100001101001101010001111001001011110001111110101000011101000001011100001001011010 0100101100011110011100110110011100111100010010011001111110101111111000000110001110000110001100001110 1110010101010001000110100011101010001010000110001111111110101010000000001111001110110101110000010011 1000010011111110001101100000101001110100011000111010011111110110111010011111010110101111011111011011
样例输出
515649254
题目来源
题意：题意说的很乱，简单来说就是输入5个长为100的‘0’ ‘1’字符串，要求从每一行中取出一个‘1’ 但是这5个‘1’得在不同的列，求有多少种取法？ 注意输出的结果乘以常数tmp=531192758
思路：我们可以用DP解决这个问题，dp[i][j]=(dp[i][j]+dp[i-1][j-1])%mod 保证当前行取得一个‘1’ 是在前一行的‘1’的后面，然后把这5行字符串用全排列颠倒顺序即可，把所有排列顺序下的dp值dp[5][100]求和即可
这题正解是状压DP复杂度O(n)=2^5*500, 我上面的DP复杂度O(n)=5!*500, 这题现场赛的时候大多数人是暴力过的，唉~ 当时我傻了，DP想了一半觉得不太可行，赛后又想了想可行，比赛的时候压力有点大，有点紧张，特别是最后半个小时。
代码如下：
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <cstring>
using namespace std;
typedef long long LL;
const LL mod=1e9+7;
char s[6][105];
LL dp[6][105],ans;
void Backtrack(int t)
{
if(t==5)
{
memset(dp,0,sizeof(dp));
for(int i=1;i<=100;i++)
{
dp[1][i]=dp[1][i-1];
if(s[1][i]==‘1‘) dp[1][i]++;
}
for(int i=2;i<=5;i++)
{
for(int j=1;j<=100;j++)
{
dp[i][j]=dp[i][j-1];
if(s[i][j]==‘1‘) dp[i][j]=(dp[i][j]+dp[i-1][j-1])%mod;
}
}
ans=(ans+dp[5][100])%mod;
return ;
}
for(int i=t;i<=5;i++)
{
swap(s[i],s[t]);
Backtrack(t+1);
swap(s[i],s[t]);
}
}
int main()
{
LL tmp=531192758; /// 常数 tmp=A(95,5)*C(90,5)*C(85,5);
while(scanf("%s",s[1]+1)!=EOF)
{
for(int i=2;i<=5;i++) scanf("%s",s[i]+1);
ans=0;
Backtrack(1);
printf("%lld\n",ans*tmp%mod);
}
return 0;
}
原文地址：https://www.cnblogs.com/chen9510/p/8323994.html