题目大意：从字符串中读取信息，然后根据公式计算
解题思路：从字符串中提取信息，并保存和标记，然后在根据公式进行计算
#include<cstdio>
#include<cmath>
#include<cstring>
double find_n(char *str, int i) {
int count = 0;
double recode[20];
double times = 1;
double ren_n = 0;
int number = 0;
int mark = 0;
double sign = 1;
int start = 2;
if(str[i+2] == '-') {
sign = -1;
start++;
}
for(int j = i + start; ; j++) {
if(str[j] >= '0' && str[j] <= '9') {
recode[count] = str[j] - '0';
count++;
if(mark == 1)
number--;
}
if(str[j] == '.')
{
mark = 1;
continue;
}
if(str[j] == 'k') {
times = 1000;
break;
}
if(str[j] == 'M') {
times = 1000000;
break;
}
if(str[j] == 'm') {
times = 0.001;
break;
}
if(str[j] == 'A' || str[j] == 'V' || str[j] == 'W')
break;
}
int j;
double k;
for( j = count - 1 , k = pow(10,number); j >= 0; j--, k = k * 10) {
ren_n = ren_n + recode[j] * k;
}
return ren_n * times * sign;
}
void put(double *num, int *mark, int count) {
double end;
if(mark[0] == 1 && (mark[1] == 2 || mark[1] == 3))
end = num[0] / num[1];
if(mark[1] == 1 && (mark[0] == 2 || mark[0] == 3))
end = num[1] / num[0];
if( (mark[1] == 2 && mark[0] == 1) || (mark[0] == 2 && mark[1] == 1))
printf("Problem #%d\nI=%.2fA\n\n", count, end);
if( (mark[1] == 3 && mark[0] == 1) || (mark[0] == 3 && mark[1] == 1))
printf("Problem #%d\nU=%.2fV\n\n", count, end);
if((mark[0] == 2 && mark[1] == 3) || (mark[1] == 2 && mark[0] == 3))
printf("Problem #%d\nP=%.2fW\n\n", count, num[1] * num[0]);
}
int main() {
int mark[2];
int number;
int send = 0;
char str[100];
double num[2];
int count ;
scanf("%d\n",&number);
for(int L= 1; L <= number; L++) {
gets(str);
count = 0;
for(int i = 0; i < strlen(str); i++) {
if(str[i + 1] == '=' && (str[i] == 'U' || str[i] == 'I' || str[i] == 'P') ) {
if(str[i] == 'U' )
mark[count] = 2;
if(str[i] == 'I' )
mark[count] = 3;
if(str[i] == 'P' )
mark[count] = 1;
num[count] = find_n(str,i -1);
count++;
}
}
put(num,mark,L);
}
return 0;
}
