PAT 1060. Are They Equal (25) 只做出19分。。

1060. Are They Equal (25)

时间限制

50 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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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提交代码

/*  太繁琐，很难全通过~~~~~~~~~~~~！！！形参变量 与 主函数变量 不要同名混淆，否则检查半天会很无语。！！！！注意程序中及时break！！  科学计数法 分为 整数 0 小数 三种情况讨论，注意补0 去0。特殊case：0021，0.0021，0，12.123*/#include<stdio.h>#include<string.h>const int MAX=200;//此处是测试点，设置为5，扣2分。。。char tmp[MAX];void init(char a2[],char b2[]){int i;for(i=0;i<MAX;i++){a2[i]=0;b2[i]=0;}}char* subStr(char m[],int p,int q,int r) //此处用(char a[],int i,int j,int w)老是出错，原来是形参变量i,j 与 主函数变量i,j 同名混淆。。{     // m[]表示一个数组，p表示开始截取的位置，q表示截取的长度，r表示整个数组的长度。         int i,k,size=0;  //由于涉及到字符串后面填充字符'0'，故需要知道 数字总位数 bool isDecimal=0;for(i=0;i<MAX;i++)tmp[i]=0;for(k=p;k<q;k++){   //如果整数部分不够有效精度，那么在后面补0.if(m[k]=='.'||k>=r){tmp[size++]='0';isDecimal=1;}else if(isDecimal==0)tmp[size++]=m[k];elsetmp[size++]='0';}return tmp;}int main(){int n,i;char a[MAX],b[MAX],a1[MAX],b1[MAX];//freopen("G:\\in.txt","r",stdin);//freopen("G:\\our.txt","w",stdout);while(scanf("%d",&n)!=EOF){init(a,b);scanf("%s%s",a,b);int len1=strlen(a), len2=strlen(b);int cnt1=0,cnt2=0;for(i=0;i<len1;i++){  //计算第1个数的整数部分总位数if(a[0]=='0')break;else if(a[i]!='.')cnt1++;else          //扫到小数点就终止扫描！！！勿忘。。。。。break;}for(i=0;i<len2;i++){  //计算第2个数的整数部分总位数if(b[0]=='0')break;else if(b[i]!='.')cnt2++;else          //扫到小数点就终止扫描！勿忘。。。。。break;}//printf("初始串：%s %s\n",a,b);strcpy(a1,subStr(a,0,n,len1));strcpy(b1,subStr(b,0,n,len2));if(cnt1==cnt2){if(strcmp(a1,b1)==0)printf("YES 0.%s*10^%d",a1,cnt1);elseprintf("NO 0.%s*10^%d 0.%s*10^%d",a1,cnt1,b1,cnt2);}else{printf("NO 0.%s*10^%d 0.%s*10^%d",a1,cnt1,b1,cnt2);}printf("\n");}return 0;}