【题解】Luogu P3123 [USACO15OPEN]贝茜说哞Bessie Goes Moo

Luogu P3123 [USACO15OPEN]贝茜说哞Bessie Goes Moo

题目描述

Farmer John and Bessie the cow love to exchange math puzzles in their free time.

The last puzzle FJ gave Bessie was quite difficult and she failed to solve it.

Now she wants to get even with FJ by giving him a challenging puzzle.

Bessie gives FJ the expression (B+E+S+S+I+E)(G+O+E+S)(M+O+O)(B+E+S+S+I+E)(G+O+E+S)(M+O+O)(B+E+S+S+I+E)(G+O+E+S)(M+O+O) , containing the

seven variables B,E,S,I,G,O,MB,E,S,I,G,O,MB,E,S,I,G,O,M (the “OOO “ is a variable, not a zero). For each

variable, she gives FJ a list of up to 500 integer values the variable can

possibly take. She asks FJ to count the number of different ways he can assign

values to the variables so the entire expression evaluates to a multiple of 7.

Note that the answer to this problem can be too large to fit into a 32-bit

integer, so you probably want to use 64-bit integers (e.g., “long long”s in C or

C++).

• 七个变量B,E,S,I,G,O,M;使得(B+E+S+S+I+E)(G+O+E+S)(M+O+O)被7整除的方案有多少种.

输入输出格式

输入格式：

The first line of the input contains an integer NNN . The next NNN lines each

contain a variable and a possible value for that variable. Each variable will

appear in this list at least once and at most 500 times. No possible value will

be listed more than once for the same variable. All possible values will be in

the range $-10^{5}$ to $10^{5}$.

输出格式：

Print a single integer, giving the number of ways FJ can assign values to

variables so the expression above evaluates to a multiple of 7.

输入输出样例

输入样例#1： 复制

10
B 2
E 5
S 7
I 10
O 16
M 19
B 3
G 1
I 9
M 2

输出样例#1： 复制

2

说明

The two possible assignments are

(B,E,S,I,G,O,M) = (2, 5, 7, 9, 1, 16, 19) -> 51,765
(B,E,S,I,G,O,M) = (2, 5, 7, 9, 1, 16, 2 ) -> 34,510

思路

• 用 $f[i][j]$表示: 第i个字母变量所代表的数mod7==j的次数
• 由于每个字母mod 7只可能有0~6共7种结果,所以直接7^7暴力枚举
• $b[k]$表示第k个变量mod 7的值 i
• $c[k]$表示输入中的数mod 7== i 的出现次数
• 当s1,s2,s3中有一个为0时,就可以累加答案

代码

#include<cmath>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#define re register int
using namespace std;
int n,x;
char a[2],s[10];  
long long ans,f[10][20],c[10],b[10];
inline int read(){
    int x=0,w=1;
    char ch=getchar();
    while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
    if(ch=='-') w=-1,ch=getchar();
    while(ch>='0'&&ch<='9') x=(x<<1)+(x<<3)+ch-48,ch=getchar();
    return x*w;
}
void dfs(int k,long long sum) {
    if(k>7) {
        int s1=(b[1]+b[2]*2+b[3]*2+b[4])%7;
        int s2=(b[5]+b[6]+b[2]+b[3])%7;
        int s3=(b[7]+2*b[6])%7;
        if(s1==0||s2==0||s3==0) ans=ans+sum;
        return;
    }
    for(re i=0;i<=6;i++) {
        b[k]=i;
        c[k]=f[k][i];
        if(c[k]==0) continue;
        dfs(k+1,sum*c[k]);
    }
}
int main() {
    //freopen("p3123.in","r",stdin);
    //freopen("p3123.out","w",stdout);
    s[1]='B';s[2]='E';s[3]='S';s[4]='I';s[5]='G';s[6]='O';s[7]='M';  
    n=read();
    for(re i=1;i<=n;i++) {
        scanf("%s%d",a,&x);
        for(re j=1;j<=7;j++)
        if(a[0]==s[j]) f[j][(x%7+7)%7]++;
    }
    dfs(1,1);     
    printf("%lld\n",ans);      
    return 0;  
}

​