ZOJ 3782 Ternary Calculation
G - Ternary Calculation
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu
Description
Complete the ternary calculation.
Input
There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:
There is a string in the form of "number1operatoranumber2operatorbnumber3". Each operator will be one of {'+', '-' , '*', '/', '%'}, and each number will be an integer in [1, 1000].
Output
For each test case, output the answer.
Sample Input
5 1 + 2 * 3 1 - 8 / 3 1 + 2 - 3 7 * 8 / 5 5 - 8 % 3
Sample Output
7 -1 0 11 3
Note
The calculation "A % B" means taking the remainder of A divided by B, and "A / B" means taking the quotient.
根据字符串等式写出结果。
我是直接暴力的。代码虽然长了点，反正也不会超时。
优化的话，可以根据运算符的优先级计算。
#include <stdio.h>
#include <string.h>
char s[3005];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int ans;
int a,b,c;
char op1,op2;
scanf("%d%*c%c%*c%d%*c%c%*c%d",&a,&op1,&b,&op2,&c);
if((op1=='+'||op1=='-')&&(op2=='+'||op2=='-'))
{
switch(op1)
{
case '+':
switch(op2)
{
case '+':
ans=a+b+c;
break;
case '-':
ans=a+b-c;
break;
}
break;
case '-':
switch(op2)
{
case '+':
ans=a-b+c;
break;
case '-':
ans=a-b-c;
break;
}
break;
}
}
if((op1=='+'||op1=='-')&&(op2=='*'||op2=='/'||op2=='%'))
{
switch(op1)
{
case '+':
switch(op2)
{
case '*':
ans=a+b*c;
break;
case '/':
ans=a+b/c;
break;
case '%':
ans=a+b%c;
break;
}
break;
case '-':
switch(op2)
{
case '*':
ans=a-b*c;
break;
case '/':
ans=a-b/c;
break;
case '%':
ans=a-b%c;
break;
}
break;
}
}
if((op1=='*'||op1=='/'||op1=='%')&&(op2=='*'||op2=='/'||op2=='%'))
{
switch(op1)
{
case '*':
switch(op2)
{
case '*':
ans=a*b*c;
break;
case '/':
ans=a*b/c;
break;
case '%':
ans=a*b%c;
break;
}
break;
case '/':
switch(op2)
{
case '*':
ans=a/b*c;
break;
case '/':
ans=a/b/c;
break;
case '%':
ans=a/b%c;
break;
}
break;
case '%':
switch(op2)
{
case '*':
ans=a%b*c;
break;
case '/':
ans=a%b/c;
break;
case '%':
ans=a%b%c;
break;
}
break;
}
}
if((op1=='*'||op1=='/'||op1=='%')&&(op2=='+'||op2=='-'))
{
switch(op1)
{
case '*':
switch(op2)
{
case '+':
ans=a*b+c;
break;
case '-':
ans=a*b-c;
break;
}
break;
case '/':
switch(op2)
{
case '+':
ans=a/b+c;
break;
case '-':
ans=a/b-c;
break;
}
break;
case '%':
switch(op2)
{
case '+':
ans=a%b+c;
break;
case '-':
ans=a%b-c;
break;
}
break;
}
}
printf("%d\n",ans);
}
return 0;
}