CSP-201903-2-24:00

24 o'clock (transfer gate)

This problem is relatively convoluted. At the beginning of violent cracking, there were a lot of steps. Later, stacks were used to solve it, which is relatively simple

Input:

10
9+3+4x3
5+4x5x5
7-9-9+8
5x6/5x4
3+5+7+9
1x1+9-9
1x9-5/9
8/5+6x9
6x7-3x6
6x4+4/5

Output:

Yes
No
No
Yes
Yes
No
No
No
Yes
Yes

Full Score code 1

#include <bits/stdc++.h>

using namespace std;

int n;
int result = 0;
bool judge[102];
string str;

int call(int a,int b,char c){
	int res;
	switch(c){
		case 'x':
			res=a*b;
			break;
		case '/':
			res=a/b;
			break;
		case '+':
			res=a+b;
			break;
		case '-':
			res=a-b;
			break;
	}
	return res;
}

bool cal(){
	char c[6];
	int coun = 0;
	int j = -1;
	c[1]=str[1];c[3]=str[3];c[5]=str[5];
	for(int i = 1; i < 6; i+=2){
		if(c[i]=='x' || c[i]=='/'){
			c[i-1] = '1';
			coun++;
		}else{
			c[i-1] = '0';
		}
	}
	if(coun == 3 || coun == 0){ //Example 1 + 2 + 3 + 4 1 * 2 * 3 * 4 1 / 2 / 3 / 4
		result = call(str[0]-'0',str[2]-'0',str[1]);
		result = call(result,str[4]-'0',str[3]);
		result = call(result,str[6]-'0',str[5]);
	}else{
		if(coun == 1){
			for(int i = 1; i < 6; i+=2){
				if(c[i-1] == '1'){
					result = call(str[i-1]-'0',str[i+1]-'0',str[i]);
					j = i;
				}
			}
			if(j == 1){			//Example 1*2+3-4
				result = call(result,str[4]-'0',str[3]);
				result = call(result,str[6]-'0',str[5]);
			}else if(j == 3){   //Example 1+2*3-4
				result = call(str[0]-'0',result,str[1]);
				result = call(result,str[6]-'0',str[5]);
			}else{              //Example 1+2+3*4
				int res = call(str[0]-'0',str[2]-'0',str[1]);
				result = call(res,result,str[3]);
			}
		}else{
			for(int i = 1; i < 6; i+=2){
				if(c[i-1] == '0'){
					j = i;
				}
			}
			if(j == 1){     //Example 1+2*3/4
				result = call(str[2]-'0',str[4]-'0',str[3]);
				result = call(result,str[6]-'0',str[5]);
				result = call(str[0]-'0',result,str[1]);
			}else if(j == 3){       //Example 1*2+3*4
				int res = call(str[0]-'0',str[2]-'0',str[1]);
				result = call(str[4]-'0',str[6]-'0',str[5]);
				result = call(res,result,str[3]);
			}else{              //Example 1*2*3-4
				result = call(str[0]-'0',str[2]-'0',str[1]);
				result = call(result,str[4]-'0',str[3]);
				result = call(result,str[6]-'0',str[5]);
			}
		}
	}
	if(result == 24){
		return true;
	}
	return false;
}

int main(){
    cin >> n;
    for(int i = 0 ; i < n; i++){
		cin >> str;
		if(cal()){
			judge[i] = true;
		}
	}
	
	for(int i = 0; i < n; i++){
		if(judge[i]){
			cout << "Yes\n";
		}else{
			cout << "No\n";
		}
	}
	
    return 0;
}

Full Score code 2 -------- using stack implementation -------- referring to other people's code to know how to use stack

#include <iostream>
using namespace std;

int main(){
	int n, num[5], temp;//Number of lines - num [top of stack, --]
	char op[5]; 	//operator
	num[0] = 0;		//Initializing the operand stack
	op[0] = 0;		//Initializing the operator stack
	cin>>n;
	int result[n];
	for(int i=0; i<n; ++i){
		for(int j=0; j<7; ++j){
			if(j%2 == 0){	//Stack of operands
				cin>>num[++num[0]];
				if (op[op[0]] == '-')	//If the top of the operator stack is - convert the operand + (- x)
					num[num[0]] *= -1;
				else if (op[op[0]] == 'x'){	//x operation required
					temp = num[num[0]] * num[num[0]-1];
					num[--num[0]] = temp;
					--op[0];
				}
				else if(op[op[0]] == '/'){	//Need / operation
					temp = num[num[0]-1] / num[num[0]];
					num[--num[0]] = temp;
					--op[0];
				}
			}
			else{			//Operator stack
				cin>>op[++op[0]];
			}
		}
		while(op[0] > 0){	//Finish the remaining + sign
				temp = num[num[0]] + num[num[0]-1];
				num[--num[0]] = temp;
				--op[0];
		}
		if(num[1] == 24)	//Decision result
			result[i] = 1;
		else
			result[i] = 0;
		num[0] = 0;			//Initializing the operand stack
	}
	
	for(int i=0; i<n; ++i){	//Print results
		if(result[i] == 1)
			cout<<"Yes\n";
		else
			cout<<"No\n";
	}
	return 0;