HDU 1427 dfs 速算24点

4个数通过 +，—，*，/和加括号，计算得24,

枚举数字和运算符，DFS即可，注意题目要求计算过程中都不能出现小数，所以做除法时稍作处理

枚举数组可用algorithm里的next_permutation

The next_permutation() function attempts to transform the given range of elements [start,end) into the next lexicographically greater permutation of elements. If it succeeds, it returns true, otherwise, it returns false.

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>

#define LL long long
int const MAX = 1e6 + 1;
int const INF = 1 << 30;
double const EPS = 0.00000001;
using namespace std;

int num[4];
bool flag;

//注意输入的10是两个字节
int getNum(char c){
if (c == 'A') return 1;
else if (c == 'J') return 11;
else if (c == 'Q') return 12;
else if (c == 'K') return 13;
else if (c == '1') return 10;
else return c - '0';
}

//括号内计算的pre,括号外的值next,计算到第pos个
void dfs(int pre, int next, int pos){
if (flag == 0) return;

//正在使用第四张
if (pos == 3){
if (pre - next == 24 || pre + next == 24 || pre * next == 24)
flag = 0;

if (next != 0 && pre % next == 0 && pre / next == 24)
flag = 0;
return;
}

//不加括号
dfs(pre + next, num[pos + 1], pos + 1);
dfs(pre - next, num[pos + 1], pos + 1);
dfs(pre * next, num[pos + 1], pos + 1);
if (next != 0 && pre % next == 0)
dfs(pre / next, num[pos + 1], pos + 1);

//加括号
dfs(pre, next + num[pos + 1], pos + 1);
dfs(pre, next - num[pos + 1], pos + 1);
dfs(pre, next * num[pos + 1], pos + 1);
if (num[pos + 1] != 0 && next % num[pos + 1] == 0)
dfs(pre, next / num[pos + 1], pos + 1);

}
int main(){
char str[3];
while (scanf("%s", str) == 1){
num[0] = getNum(str[0]);
for (int i = 1; i < 4; i++){
scanf("%s", str);
num[i] = getNum(str[0]);
}
flag = 1;
sort(num, num + 4);
do {
dfs(num[0], num[1], 1);
} while (flag && next_permutation(num, num + 4));

printf("%s\n", flag ? "No" : "Yes");
}
return 0;
}

/*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 99999999
using 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]);

7fe0
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;
}