hdu1427之速算24点

速算24点

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2430    Accepted Submission(s): 582

Problem Description

速算24点相信绝大多数人都玩过。就是随机给你四张牌，包括A(1),2,3,4,5,6,7,8,9,10,J(11),Q(12),K(13)。要求只用'+','-','*','/'运算符以及括号改变运算顺序，使得最终运算结果为24(每个数必须且仅能用一次)。游戏很简单，但遇到无解的情况往往让人很郁闷。你的任务就是针对每一组随机产生的四张牌，判断是否有解。我们另外规定，整个计算过程中都不能出现小数。

 

Input

每组输入数据占一行，给定四张牌。

 

Output

每一组输入数据对应一行输出。如果有解则输出"Yes"，无解则输出"No"。

 

Sample Input

A 2 3 63 3 8 8

 

Sample Output

YesNo

 

/*分析:对于a,b,c,d四个数进行+,-,*,/。
有这几种情况:1.(a@b)@(c@d),(c@d)@(a@b),(b@a)@(c@d),... 综合为(x1@y1)@(x1@y1)这种情况,不过得到的值可以为-24 
2.((a@b)@c)@d,a@((b@c)@d),(d@(a@b)@c),...综合为((x1@y1)@x2)@y2这种情况,不过得到的值可以为-24 
*/ 

所以只要求a,b,c,d的全排列在用(a@b)@(c@b),((a@b)@c))@d去计算即可

1.用next_permutation求全排列:注意next_permutation求的是按字典序的全排列，所以要先排序

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<queue>#include<algorithm>#include<map>#include<iomanip>#define INF 99999999using namespace std;const int MAX=5;char s[3];int number[MAX];void check(char ch,int &num){if(ch == 'A')num=1;else if(ch == 'J')num=11;else if(ch == 'Q')num=12;else if(ch == 'K')num=13;else for(int i=2;i<10;++i){if(ch == i+'0'){num=i;break;}}if(ch == '1')num=10;}int f(int a,int op,int c){if(op == 0)return a+c;if(op == 1)return a-c;if(op == 2)return a*c;if(c && a%c == 0)return a/c;return INF;}bool calculate(int i,int j,int k){int temp1,temp2;temp1=f(number[0],i,number[1]);if(temp1 != INF)temp2=f(number[2],k,number[3]);if(temp2 == INF)temp1=INF;if(temp1 != INF)temp1=f(temp1,j,temp2);if(temp1 == 24 || temp1 == -24)return true;temp1=f(number[0],i,number[1]);if(temp1 != INF)temp1=f(temp1,j,number[2]);if(temp1 != INF)temp1=f(temp1,k,number[3]);if(temp1 == 24 || temp1 == -24)return true;return false;}int main(){while(~scanf("%s",s)){check(s[0],number[0]);for(int i=1;i<=3;++i){scanf("%s",s);check(s[0],number[i]);}sort(number,number+4);bool flag=false;do{for(int i=0;i<4 && !flag;++i)for(int j=0;j<4 && !flag;++j)for(int k=0;k<4 && !flag;++k){flag=calculate(i,j,k);//if(flag)cout<<number[0]<<' '<<number[1]<<' '<<number[2]<<' '<<number[3]<<endl;//if(flag)cout<<i<<' '<<j<<' '<<k<<endl;}}while(!flag && next_permutation(number,number+4));if(flag)printf("Yes\n");else printf("No\n");}return 0;}

自己写的全排列:

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<queue>#include<algorithm>#include<map>#include<iomanip>#define INF 99999999using namespace std;const int MAX=5;char s[3];int number[MAX];bool flag;void check(char ch,int &num){if(ch == 'A')num=1;else if(ch == 'J')num=11;else if(ch == 'Q')num=12;else if(ch == 'K')num=13;else for(int i=2;i<10;++i){if(ch == i+'0'){num=i;break;}}if(ch == '1')num=10;}int f(int a,int op,int c){if(op == 0)return a+c;if(op == 1)return a-c;if(op == 2)return a*c;if(c && a%c == 0)return a/c;return INF;}bool calculate(int i,int j,int k){int temp1,temp2;temp1=f(number[0],i,number[1]);if(temp1 != INF)temp2=f(number[2],k,number[3]);if(temp2 == INF)temp1=INF;if(temp1 != INF)temp1=f(temp1,j,temp2);if(temp1 == 24 || temp1 == -24)return true;temp1=f(number[0],i,number[1]);if(temp1 != INF)temp1=f(temp1,j,number[2]);if(temp1 != INF)temp1=f(temp1,k,number[3]);if(temp1 == 24 || temp1 == -24)return true;return false;}void Perm(int k){if(k == 3){for(int i=0;i<4 && !flag;++i)for(int j=0;j<4 && !flag;++j)for(int k=0;k<4 && !flag;++k){flag=calculate(i,j,k);}return;}Perm(k+1);for(int i=k+1;i<4;++i){if(flag)return;if(number[i] == number[k])continue;swap(number[k],number[i]);Perm(k+1);swap(number[k],number[i]);}}int main(){while(~scanf("%s",s)){check(s[0],number[0]);for(int i=1;i<=3;++i){scanf("%s",s);check(s[0],number[i]);}flag=false;Perm(0);if(flag)printf("Yes\n");else printf("No\n");}return 0;}