Pat1060代码
题目描述:
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
AC代码:
浮点数的科学计数法,纯字符串的比较;
下面是几个测试用例:
1.有前导0的情况 例如:
3 00000012.3 0000001.1
NO 0.123*10^2 0.11*10^1(太坑,一点提示也没有)
2.还有就是当位数不足N位时,不需要补全 , 如上的 0000001.1
3.0的不同表示,如:
5 0000000 0000000.000000
YES 0.00000*10^0
其实这题有些搞笑, 如下:
4 00000012.0 0000012.00
NO 0.120*10^2 0.1200*10^2
同样是0.12不相等。。。 纯字符串比较,题目要求不删除小数部分的后导0
#include<cstdio> #include<cstring> #define MAX 105 using namespace std; int Process(char str[],int n,char ret[]) { int i=0,j=0; int exp=0; while(str[i]==‘0‘)//去掉前面的0 i++; if(str[i]==‘.‘)//型如0000000.000*****的浮点数 { i++; while(str[i]==‘0‘) { i++; exp--; } if(str[i]==‘\0‘)//0的另一种形式如00000.000000 { int k; exp=0; for(k=0;k<n;k++) ret[k]=‘0‘; ret[k]=‘\0‘; return exp; } while(str[i]!=‘\0‘) { ret[j]=str[i]; i++; j++; if(j==n) break; } } else { if(str[i]==‘\0‘)//0的一种表示例如00000000 { int k; exp=0; for(k=0;k<n;k++) ret[k]=‘0‘; ret[k]=‘\0‘; return exp; } else//0000*****.****** { while(str[i]!=‘.‘&&str[i]!=‘\0‘) { if(j<n)//取n位的精度 { ret[j]=str[i]; j++; } i++; exp++; } if(j<n&&str[i]!=‘\0‘) { i++; while(str[i]!=‘\0‘) { ret[j]=str[i]; i++; j++; if(j==n) break; } } } } ret[j]=‘\0‘; return exp; } int main(int argc,char *argv[]) { int N; char A[MAX],B[MAX]; scanf("%d %s %s",&N,A,B); char retA[MAX],retB[MAX]; int expA=Process(A,N,retA); int expB=Process(B,N,retB); if(expA==expB&&strcmp(retA,retB)==0) printf("YES 0.%s*10^%d\n",retA,expA); else printf("NO 0.%s*10^%d 0.%s*10^%d\n",retA,expA,retB,expB); return 0; }
Pat(Advanced Level)Practice--1060(Are They Equal)