http://pat.zju.edu.cn/contests/pat-a-practise/1060

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

下面的代码为嘛过不了呢=，=

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
using namespace std;
struct ans{
	string d;
	int k;
};
ans convert(string s, int n){
	ans tmp;
	if ((s[0] != '0') && (s.find(".") != string::npos)){
		tmp.k = s.find(".");
		if (s.size() - 1 >= n){
			if (tmp.k >= n)
				tmp.d = s.substr(0, n);
			else if (tmp.k < n){
				tmp.d = s.substr(0, tmp.k) + s.substr(tmp.k+1, n-tmp.k);
			}
		}
		else{
			tmp.d = s.substr(0, tmp.k) + s.substr(tmp.k + 1, s.size()-1-tmp.k);
			for (int i = 0; i < n - (s.size() - 1); i++){
				tmp.d += "0";
			}
		}
	}
	if ((s[0] != '0') && (s.find(".") == string::npos)){
		tmp.k = s.size();
		if (n <= tmp.k)
			tmp.d = s.substr(0, n);
		else if (n > tmp.k){
			tmp.d = s;
			for (int i = 0; i < n - tmp.k; i++){
				tmp.d += "0";
			}
		}
	}
	if (s[0] = '0' && s != "0"){
		int t;
		for (int i = 2; i < s.size(); i++){
			if (s[i] != '0'){
				t = i;
				tmp.k = -(i - 1);
				break;
			}
		}
		if (s.size() - t >= n)
			tmp.d = s.substr(t, n);
		else
		{
			tmp.d = s.substr(t, n);
			for (int i = 0; i < n - s.size() + t; i++){
				tmp.d += "0";
			}
		}
	}
	if (s == "0"){
		tmp.k = 0;
		tmp.d = "";
		for (int i = 0; i < n; i++){
			tmp.d += "0";
		}
	}
	return tmp;
}
int main(){
	int n;
	string s1, s2;
	ans tmp1, tmp2;
	cin >> n >> s1 >> s2;
	tmp1 = convert(s1, n);
	tmp2 = convert(s2, n);
	if (tmp1.d == tmp2.d && tmp1.k == tmp2.k)
		cout << "YES" << " " << "0." << tmp1.d << "*10^" << tmp1.k << endl;
	else
		cout << "NO" << " " << "0." << tmp1.d << "*10^" << tmp1.k << " " << "0." << tmp2.d << "*10^" << tmp2.k << endl;
	return 0;
}

我只能说坑太多，各种格式的数字也没描述清楚。。。真操蛋

AC代码： 

#include <stdio.h>
#include <string.h>
#define MAX 110
struct result{
	char d[MAX];
	int k;
};

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)
			continue;
		else{
			if (firstPos == -1)
				firstPos = i;
			if (index < n)
			{
				if (index < strlen(a))
					r.d[index++] = a[i];
				else
					r.d[index++] = '0';
			}
		}
	}
	r.d[index] = 0;
	if (pointPos == -1)
		pointPos = i;
	if (pointPos - firstPos < 0)
		r.k = pointPos - firstPos + 1;
	else
		r.k = pointPos - firstPos;
	if (index == 0){
		int 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);
	else
		printf("NO 0.%s*10^%d 0.%s*10^%d\n", r1.d, r1.k, r2.d, r2.k);
	return 0;
}