1060. Are They Equal (25)【字符串處理】——PAT (Advanced Level) Practise

1060. Are They Equal (25)【字符串處理】——PAT (Advanced Level) Practise

題目信息

1060. Are They Equal (25)

時間限制50 ms
內存限制65536 kB
代碼長度限制16000 B

If a machine can save only 3 significant digits, the float numbers 12300 and 12358.9 are considered equal since they are both saved as 0.123*10^5 with simple chopping. Now given the number of significant digits on a machine and two float numbers, you are supposed to tell if they are treated equal in that machine.

Input Specification:

Each input file contains one test case which gives three numbers N, A and B, where N (<100) is the number of significant digits, and A and B are the two float numbers to be compared. Each float number is non-negative, no greater than 10100, and that its total digit number is less than 100.

Output Specification:

For each test case, print in a line “YES” if the two numbers are treated equal, and then the number in the standard form “0.d1…dN*10^k” (d1>0 unless the number is 0); or “NO” if they are not treated equal, and then the two numbers in their standard form. All the terms must be separated by a space, with no extra space at the end of a line.

Note: Simple chopping is assumed without rounding.

Sample Input 1:
3 12300 12358.9
Sample Output 1:
YES 0.123*10^5
Sample Input 2:
3 120 128
Sample Output 2:
NO 0.120*10^3 0.128*10^3

解題思路

記錄整數部分長度，並刪掉小數點，去掉小數部分末尾的0，轉成結果格式比較是否相等

AC代碼

#include 
#include 
int tran(char *s, char *r, int n){
    char *p, *pr;
    int len, fr = 0, t = 0;
    if (p = strchr(s, '.')){ 
        char *tp = s + strlen(s) - 1;
        while (tp != p && *tp == '0'){ //清理小數部分末尾0
            *tp = '\0';
            --tp;
        }
        len = p - s;
        while (*p){
            *p = *(p+1);
            ++p;
        }
    }else{
        len = strlen(s);
    }
    p = s;
    while (*p){
        if (*p == '0'){
            ++fr;
        }else{
            break;
        }
        ++p;
    }
    strcpy(r, "0.");
    pr = &r[2];
    while (*p && t < n){
        *pr++ = *p++;
        ++t;
    }
    while (t++ < n){
        *pr++ = '0';
    }
    sprintf(pr, "*10^%d", len - fr);
    return len;
}
int main()
{
    int n;
    char s1[200], s2[200], rs1[200], rs2[200];
    scanf("%d%s%s", &n, s1, s2);
    int len1 = tran(s1, rs1, n);
    int len2 = tran(s2, rs2, n);
    if (0 == strcmp(rs1, rs2)){
        printf("YES %s\n", rs1);
    }else{
        printf("NO %s %s\n", rs1, rs2);
    }
    return 0;
}