PAT (Advanced Level) Practise 1060 Are They Equal (25)

1060. Are They Equal (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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*10⁵ 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...d_N*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

题意：给你两个数，将它们用小数点后k位的科学计数法表示，比较一下两个是否一样

解题思路：这题会有前导零的数据，所以关键在于找到第一个有效数字的位置和小数点的位置，然后就可以处理出一个数字小数点后几位的科学计数法的表示方式

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int n,x1,x2;char s1[200], s2[200], ans1[200], ans2[200];void check(char ch1[], char ch2[],int &x){int len = strlen(ch1);int k = len, kk;for(int i=0;i<len;i++)if (ch1[i] == '.') { k = i; break; }kk = -1;for(int i=0;i<k;i++)if (ch1[i] != '0') { kk = i; break; }if (kk == -1){int p = -1;for(int i=k+1;i<len;i++)if (ch1[i] != '0') { p = i; break; }if (p == -1){ch2[0] = '0', ch2[1] = '.';for (int i = 1; i <= n; i++) ch2[i + 1] = '0';ch2[n + 2] = '\0';x = 0;}else{x = -(p - k - 1);ch2[0] = '0', ch2[1] = '.';int cnt = 2;while (p < len&&cnt < n + 2) ch2[cnt++] = ch1[p], p++;while (cnt < n + 2) ch2[cnt++] = '0';ch2[cnt] = '\0';}}else{x = k-kk;ch2[0] = '0', ch2[1] = '.';int cnt = 2;while (ch1[kk] != '.'&&kk < len&&cnt < n + 2) ch2[cnt++] = ch1[kk],kk++;kk++;while(kk<len&&cnt<n+2) ch2[cnt++] = ch1[kk], kk++;while (cnt < n + 2) ch2[cnt++] = '0';ch2[cnt] = '\0';}}int main(){while (~scanf("%d%s%s", &n,s1,s2)){check(s1, ans1,x1);check(s2, ans2,x2);if (!strcmp(ans1, ans2) && x1 == x2) printf("YES %s*10^%d\n", ans1, x1);else printf("NO %s*10^%d %s*10^%d\n", ans1, x1,ans2,x2);}return 0;}