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

本文探讨了在给定的机器上如何通过简单裁剪的方式比较两个浮点数,并判断它们是否被视为相等。包括输入参数的解析、整数部分位数的计算、科学计数法的处理以及输出格式的实现。

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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...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

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];
		else
			tmp[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);
			else
				printf("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;
}

 

(c++题解,代码运行时间小于200ms,内存小于64MB,代码不能有注释)It’s time for the company’s annual gala! To reward employees for their hard work over the past year, PAT Company has decided to hold a lucky draw. Each employee receives one or more tickets, each of which has a unique integer printed on it. During the lucky draw, the host will perform one of the following actions: Announce a lucky number x, and the winner is then the smallest number that is greater than or equal to x. Ask a specific employee for all his/her tickets that have already won. Declare that the ticket with a specific number x wins. A ticket can win multiple times. Your job is to help the host determine the outcome of each action. Input Specification: The first line contains a positive integer N (1≤N≤10 5 ), representing the number of tickets. The next N lines each contains two parts separated by a space: an employee ID in the format PAT followed by a six-digit number (e.g., PAT202412) and an integer x (−10 9 ≤x≤10 9 ), representing the number on the ticket. Then the following line contains a positive integer Q (1≤Q≤10 5 ), representing the number of actions. The next Q lines each contain one of the following three actions: 1 x: Declare the ticket with the smallest number that is greater than or equal to x as the winner. 2 y: Ask the employee with ID y all his/her tickets that have already won. 3 x: Declare the ticket with number x as the winner. It is guaranteed that there are no more than 100 actions of the 2nd type (2 y). Output Specification: For actions of type 1 and 3, output the employee ID holding the winning ticket. If no valid ticket exists, output ERROR. For actions of type 2, if the employee ID y does not exist, output ERROR. Otherwise, output all winning ticket numbers held by this employee in the same order of input. If no ticket wins, output an empty line instead. Sample Input: 10 PAT000001 1 PAT000003 5 PAT000002 4 PAT000010 20 PAT000001 2 PAT000008 7 PAT000010 18 PAT000003 -5 PAT102030 -2000 PAT000008 15 11 1 10 2 PAT000008 2 PAT000001 3 -10 1 9999 1 -10 3 2 1 0 3 1 2 PAT000001 3 -2000 Sample Output: PAT000008 15 ERROR ERROR PAT000003 PAT000001 PAT000001 PAT000001 1 2 PAT102030
最新发布
08-12
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