题解 | 特殊的科学计数法

#include <stdio.h>
#include <string.h>
int main() {
    char n[100000];
    scanf("%s", n);
    int len = strlen(n);
    int c = len - 1;
    int a = n[0] - '0';
    int b = n[1] - '0';
    int d = n[2] - '0';
    float e = a*10.0+b+d/10.0;
    int E = d>=5?(int)e+1:(int)e;
    if(E<100){
        a = E/10;
        b = E%10;
    }else{
        a = 1;
        b = 0;
        c++;
    }

    printf("%d.%d*10^%d", a, b, c);
    return 0;
}