PAT [A1060]-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.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

解题思路：
字符串操作，注意0.0， 0.001等数据
sprintf可以将数字传进对应的char数组，然后再转入string中

AC代码：

#include <cstdio>#include <cstring>#include <string>#include <iostream>using namespace std;const int maxn = 110;int N;string A, B;string trans(int N, string dat){    string ans = "", sub = "";    int pPos = dat.length(), nPos = -1, lenT = dat.length();    // 定位小数点与第一个非0数字的位置    for (int i = 0; i < dat.length(); i++){        if (dat[i] == '.') {            pPos = i;            break;        }    }    for (int i = 0; i < dat.length(); i++){        if (dat[i] != '0' && dat[i] != '.') {            nPos = i;            break;        }    }    // 若位数目不够，则补0    if (lenT - nPos - 1 < N){        for (int i = 0; i < (N - lenT - nPos + 1); i++){            dat += "0";        }    }    // 输入为0    if (nPos == -1) {        for (int i = 0; i < N; i++){            sub += "0";        }        ans = "0." + sub + "*10^0";    }    // 输入不为0    else{        int len = pPos - nPos;        // 如果len>0，代表整数部分不为0        if (len > 0){            // 若无小数点或截取部分没有超出整数部分，则只截取整数部分, 否则还需要截取小数部分            if (pPos == lenT || N <= len) sub = dat.substr(nPos, N);            else{                sub = dat.substr(nPos, len) + dat.substr(pPos + 1, N - len);            }        }        // 如果len<0，代表整数部分为0，应为10的（len+1）次方        else{            len += 1;            sub = dat.substr(nPos, N);        }        // 将数组传进string        char lenS[maxn];        sprintf(lenS, "%d", len);        string lenStr = lenS;        ans = "0." + sub + "*10^" + lenS;    }    return ans;}int main(){    freopen("C:\\Users\\Administrator\\Desktop\\test.txt", "r", stdin);    while (scanf("%d", &N) != EOF){        cin >> A >> B;        //cout << A << " "<< B << endl;        string ansA = trans(N, A), ansB = trans(N, B);        if (ansA == ansB) cout << "YES " << ansA << endl;        else cout << "NO " << ansA << " " << ansB << endl;    }    fclose(stdin);    return 0;}