1060. Are They Equal (25)

**转自https://www.liuchuo.NET/archives/2293**

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

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题目大意：给出两个数，问将它们写成保留N位小数的科学计数法后是否相等。如果相等，输出YES，输出他们的科学记数法表示方法。如果不相等输出NO，分别输出他们的科学计数法
分析：
1. cnta 和 cntb 通过扫描字符串得到小数点所在的下标（初始化cnta cntb为字符串长度，即下标为strlen(str））
2. 考虑到可能前面有多余的零，用 p 和 q 通过扫描字符串使 p q 开始于第一个非0（且非小数点）处的下标
3. 如果cnta >= p ，说明小数点在第一个开始的非0数的下标的右边，那么科学计数法的指数为cnta – p ; 否则应该为cnta – p + 1; 字符串b同理。
4. 如果 p 和 q 等于字符串长度， 说明字符串是 0， 此时直接把 cnta（或者cntb）置为0，因为对于0来说乘以几次方都是相等的，如果不置为0可能会出现两个0比较导致判断为它们不相等
5. indexa = 0开始给新的A数组赋值，共赋值n位除去小数点外的正常数字，从p的下标开始。如果p大于等于strlen，说明字符串遍历完毕后依旧没能满足需要的位数，此时需要在A数组后面补上0直到满足n位数字。indexb同理，产生新的B数组
6. 判断A和B是否相等，且cnta和cntb是否相等。如果相等，说明他们用科学计数法表示后是相同的，输出YES，否则输出NO，同时输出正确的科学计数法
注意：
– 10的0次方和1次方都要写。
– 题目中说，无需四舍五入。
– 数组开大点，虽然只有100位，但是很有可能前面的0很多导致根本不止100位。一开始开的110，几乎没有AC的任何测试点。。后来开了10000就AC了~

#include<iostream>#include<cstring>#include<cstdio>using namespace std;int get_cnt(char a[]){    for (int i=0;i<strlen(a);i++)    {        if (a[i]=='.')            return i;    }    return strlen(a);}int main(){    int n,p=0,q=0;    char a[10000],b[10000],A[10000],B[10000];    scanf("%d %s %s",&n,a,b);    int cnta=get_cnt(a);    int cntb=get_cnt(b);    while(a[p]=='0'||a[p]=='.') p++;    while(b[q]=='0'||b[q]=='.') q++;    cnta=cnta>=p?cnta-p:cnta-p+1;    cntb=cntb>=q?cntb-q:cntb-q+1;    if (p==strlen(a)) cnta=0;    if (q==strlen(b)) cntb=0;    int indexa=0,indexb=0;    while(indexa<n)    {        if (a[p]!='.'&&p<strlen(a))            A[indexa++]=a[p];        else if (p>=strlen(a))            A[indexa++]='0';        p++;    }    while(indexb<n)    {        if (b[q]!='.'&&q<strlen(b))            B[indexb++]=b[q];        else if(q>=strlen(b))            B[indexb++]='0';        q++;    }    if (strcmp(A,B)==0&&cnta==cntb)        printf("YES 0.%s*10^%d",A,cnta);    else        printf("NO 0.%s*10^%d 0.%s*10^%d",A,cnta,B,cntb);;    return 0;}