1060 Are They Equal

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.d[1]...d[N]*10^k (d[1]>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

解析：

然后设计一个函数，对于输入的数字（存储在char*内）数组进行遍历。
遍历时定义firstPos记录第一个非0数，pointPos记录小数点位置。

需要注意以下问题：

①题目的case似乎出现了00123.45这种坑爹的情况，因此要处理无效的0，也就说开头出现的0是不能要的，只有碰到第一个非0才记录firstPos。

②对于题目给定的精度，如果输入的数字位数不够，要补0，在结尾处还要补\0使得字符串可靠结束。

③处理完毕后，对于firstPos和pointPos做比较，如果前者小，说明是大于1的数，pointPos - firstPos即为10的几次方；如果后者小，说明是小数，应该用firstPos - pointPos + 1，+1是因为firstPos越过了小数点，而小数点不能算作一位，注意的是这个次方是负值。

④比较相等时，如果基数部分一致、次方也一致，才算相等。

代码如下：

#include<bits/stdc++.h>
using namespace std;#define e exp(1)
#define pi acos(-1)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define ll long long
#define ull unsigned long long
#define mem(a,b) memset(a,b,sizeof(a))
int gcd(int a,int b){return b?gcd(b,a%b):a;}const int MAX=110;
struct result{char d[MAX]; // 0.xxx部分int k; // 10的k次方
};result getResult(char *a, int n){result r;int firstPos = -1;int pointPos = -1;int index = 0;int i;for (i = 0; a[i]; i++){if (a[i] == '.'){pointPos = i;continue;}else if (a[i] == '0' && firstPos == -1) // 不能以0开头，否则忽略continue;else{if (firstPos == -1)firstPos = i; // 第一个非0数字的位置if (index < n){if (index < strlen(a))r.d[index++] = a[i]; // 对于特定的精度，有数字则填入相应数字，没有则补0elser.d[index++] = '0';}}}r.d[index] = 0; // 在数字结尾加\0，防止越界if (pointPos == -1)pointPos = i; // 如果没有找到小数点，则小数点在最后，这是个纯整数if (pointPos - firstPos < 0) // 判断小数点与第一个非0数字的位置关系，计算10的几次方r.k = - (firstPos - pointPos - 1); // 负次方，例如0.015，pointPos = 1, firstPos = 3, 3 - 1 - 1 = 1, -1是因为多算了小数点进去,0.15*10^-1elser.k = pointPos - firstPos; // 正次方，例如21.25，pointPos = 2，firstPos = 0，2-0=2，0.2125*10^2if (index == 0){ // 如果index = 0，代表值为0，则每一位都写0，再加\0int i;for (i = 0; i != n; i++)r.d[i] = '0';r.d[i] = 0;r.k = 0;}return r;
}int main(){int n;char a[MAX], b[MAX];scanf("%d%s%s", &n, a, b);result r1 = getResult(a, n);result r2 = getResult(b, n);if (strcmp(r1.d, r2.d) == 0 && r1.k == r2.k)printf("YES 0.%s*10^%d\n", r1.d, r1.k);elseprintf("NO 0.%s*10^%d 0.%s*10^%d\n", r1.d, r1.k, r2.d, r2.k);return 0;
}