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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.
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.
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.
3 12300 12358.9
YES 0.123*10^5
3 120 128
NO 0.120*10^3 0.128*10^3
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
英文 | 解释 |
---|---|
chopping | 碎片 |
原文链接:https://blog.csdn.net/liuchuo/article/details/52262703
乙级中的一道科学计数法的题目 https://blog.csdn.net/qq_41581765/article/details/119791168 对比学习~
#include <iostream>
#include <cstring>
using namespace std;
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 = strlen(a), cntb = strlen(b);
for(int i = 0; i < strlen(a); i++) {
if(a[i] == '.') {
cnta = i;
break;
}
}
for(int i = 0; i < strlen(b); i++) {
if(b[i] == '.') {
cntb = i;
break;
}
}
int indexa = 0, indexb = 0;
while(a[p] == '0' || a[p] == '.') p++;
while(b[q] == '0' || b[q] == '.') q++;
if(cnta >= p)
cnta = cnta - p;
else
cnta = cnta - p + 1;
if(cntb >= q)
cntb = cntb - q;
else
cntb = cntb - q + 1;
if(p == strlen(a))
cnta = 0;
if(q == strlen(b))
cntb = 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;
}
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