DD11. 幂运算
描述
输入描述
多组测试用例，请参考例题的输入处理 输入每行一个浮点数 R 其中0.0 < R <99.999， 一个整数 n 其中0 < n <=25输出描述
输出R的n次方示例1
输入:
95.123 12 0.1 1
输出:
548815620517731830194541.899025343415715973535967221869852721 0.1
C 解法, 执行用时: 2ms, 内存消耗: 232KB, 提交时间: 2018-09-18
#include <stdio.h>
#include <string.h>
int len; //total length of exponentiation result.
int product[126] = {0}; // storing result, at most length 5*25 + 1 = 126
void multiply(int a[], int n)
{
int i;
int carry = 0; // a carry number in multiplying
for (i = 0; i < len; i++)
{
int temp = a[i]*n + carry;
a[i] = temp % 10;
carry = temp / 10;
}
while (carry)
{
a[i++] = carry % 10;
carry /= 10;
}
len = i;
}
int main(int argc, char* argv[])
{
int n; // power n
char s[6]; // real number R, at most the length is 6
while (scanf("%s %d", s, &n) != EOF)
{
int position=0, i=0, num=0, j=0;
for (i=0; i<strlen(s); i++)
{
if (s[i] == '.')
{
position = (strlen(s) - 1 - i) * n; // calculate decimal point position after R^n
}
else
{
num = num*10 + s[i] - 48; // transfer float to integer
}
}
// product calculation
product[0]=1;
len = 1;
for (i = 0; i < n; i++)
{
multiply(product, num);
}
// format output
if (len <= position) // product is less than 1
{
printf("0");
printf("."); // print decimal point
for (i=0; i<position-len; i++)
{
printf("0"); // print zero between decimal point and decimal
}
j = 0;
//while (product[j] == 0) // trim trailing zeros
//{
// j++;
//}
for (i=len-1; i>=j; i--)
{
printf("%d", product[i]);
}
}
else
{
j=0;
while (product[j]==0 && j<position) // trim trailing zeros
{
j++;
}
for (i=len-1; i>=j; i--)
{
if (i+1 == position) // cause index in C language starts from 0
{
printf(".");
}
printf("%d", product[i]);
}
}
}
}
C 解法, 执行用时: 2ms, 内存消耗: 328KB, 提交时间: 2019-05-13
#include <stdio.h>
#include <string.h>
int len; //total length of exponentiation result.
int product[126] = {0}; // storing result, at most length 5*25 + 1 = 126
void multiply(int a[], int n)
{
int i;
int carry = 0; // a carry number in multiplying
for (i = 0; i < len; i++)
{
int temp = a[i]*n + carry;
a[i] = temp % 10;
carry = temp / 10;
}
while (carry)
{
a[i++] = carry % 10;
carry /= 10;
}
len = i;
}
int main(int argc, char* argv[])
{
int n; // power n
char s[6]; // real number R, at most the length is 6
while (scanf("%s %d", s, &n) != EOF)
{
int position=0, i=0, num=0, j=0;
for (i=0; i<strlen(s); i++)
{
if (s[i] == '.')
{
position = (strlen(s) - 1 - i) * n; // calculate decimal point position after R^n
}
else
{
num = num*10 + s[i] - 48; // transfer float to integer
}
}
// product calculation
product[0]=1;
len = 1;
for (i = 0; i < n; i++)
{
multiply(product, num);
}
// format output
if (len <= position) // product is less than 1
{
printf("0");
printf("."); // print decimal point
for (i=0; i<position-len; i++)
{
printf("0"); // print zero between decimal point and decimal
}
j = 0;
//while (product[j] == 0) // trim trailing zeros
//{
// j++;
//}
for (i=len-1; i>=j; i--)
{
printf("%d", product[i]);
}
}
else
{
j=0;
while (product[j]==0 && j<position) // trim trailing zeros
{
j++;
}
for (i=len-1; i>=j; i--)
{
if (i+1 == position) // cause index in C language starts from 0
{
printf(".");
}
printf("%d", product[i]);
}
}
}
}
C 解法, 执行用时: 2ms, 内存消耗: 352KB, 提交时间: 2018-09-07
#include <stdio.h>
#include <string.h>
int len; //total length of exponentiation result.
int product[126] = {0}; // storing result, at most length 5*25 + 1 = 126
void multiply(int a[], int n)
{
int i;
int carry = 0; // a carry number in multiplying
for (i = 0; i < len; i++)
{
int temp = a[i]*n + carry;
a[i] = temp % 10;
carry = temp / 10;
}
while (carry)
{
a[i++] = carry % 10;
carry /= 10;
}
len = i;
}
int main(int argc, char* argv[])
{
int n; // power n
char s[6]; // real number R, at most the length is 6
while (scanf("%s %d", s, &n) != EOF)
{
int position=0, i=0, num=0, j=0;
for (i=0; i<strlen(s); i++)
{
if (s[i] == '.')
{
position = (strlen(s) - 1 - i) * n; // calculate decimal point position after R^n
}
else
{
num = num*10 + s[i] - 48; // transfer float to integer
}
}
// product calculation
product[0]=1;
len = 1;
for (i = 0; i < n; i++)
{
multiply(product, num);
}
// format output
if (len <= position) // product is less than 1
{
printf("0");
printf("."); // print decimal point
for (i=0; i<position-len; i++)
{
printf("0"); // print zero between decimal point and decimal
}
j = 0;
//while (product[j] == 0) // trim trailing zeros
//{
// j++;
//}
for (i=len-1; i>=j; i--)
{
printf("%d", product[i]);
}
}
else
{
j=0;
while (product[j]==0 && j<position) // trim trailing zeros
{
j++;
}
for (i=len-1; i>=j; i--)
{
if (i+1 == position) // cause index in C language starts from 0
{
printf(".");
}
printf("%d", product[i]);
}
}
printf("\n");
}
}
C++ 解法, 执行用时: 2ms, 内存消耗: 352KB, 提交时间: 2018-09-02
#include <iostream>
#include <cstring>
using namespace std;
int getSmallNumBit(string &num){
int count = 0;
for(int i = 0; i < num.length(); i++){
if(num[i] == '.'){
count = num.length() - i -1;
while(i < num.length()-1){
num[i] = num[i+1];
i++;
}
}
}
if(count > 0)
num.pop_back();
return count;
}
string multi(string a, string b){
int smallNumLen = getSmallNumBit(a) + getSmallNumBit(b);
int res[1000];
int k = 0;//指示每次操作的起点
int lena = a.length();
int lenb = b.length();
memset(res,0, sizeof(res));
int t = 0;//进位
for(int i = lena - 1; i >= 0; i--){//乘数
t = 0;
k = lena -1 - i;
for(int j = lenb - 1; j >= 0; j--){
res[k] = res[k] + (a[i]-'0') * (b[j] - '0')+ t;
t = res[k] / 10;
res[k] = res[k] % 10;
k++;
}
if(t > 0){
res[k] = t;
}
}
if (res[k] == 0){
k--;
}
int start = 0;
int end = k;
while(start<end){
swap(res[start],res[end]);
end--;
start++;
}
end = k;
char resChar[1000];
memset(resChar,0,sizeof(resChar));
int loc = 0;
if(smallNumLen > 0){
while(end>=0){
if(loc == smallNumLen){
res[end+1] = '.' - '0';
break;
}else{
res[end+1] = res[end];
end--;
}
loc++;
}
k++;
}
for(int i = 0; i <=k ; i++){
resChar[i] = (char)(res[i] + '0');
}
return string(resChar);
}
void delEnd0(string &num){
if(num[num.length()-1] != '0') return;
for(int i = num.length()-1; i >= 0; i--){
if(num[i] == '0'){
num.pop_back();
}else{
break;
}
}
}
int main(){
// string s = multi("512","2");
// delEnd0(s);
// cout<<s<<endl;
//freopen("/Users/xiaojian/CLionProjects/Bishi/in.txt","r",stdin);
int n;
string R;
cin>>R>>n;
string res = R;
if(n==1){
cout<<res<<endl;
return 0;
}
for(int i = 1; i < n; i++){
res = multi(res,R);
}
delEnd0(res);
cout<<res<<endl;
return 0;
}
C 解法, 执行用时: 2ms, 内存消耗: 364KB, 提交时间: 2019-03-31
#include <stdio.h>
#include <string.h>
int len; //total length of exponentiation result.
int product[126] = {0}; // storing result, at most length 5*25 + 1 = 126
void multiply(int a[], int n)
{
int i;
int carry = 0; // a carry number in multiplying
for (i = 0; i < len; i++)
{
int temp = a[i]*n + carry;
a[i] = temp % 10;
carry = temp / 10;
}
while (carry)
{
a[i++] = carry % 10;
carry /= 10;
}
len = i;
}
int main(int argc, char* argv[])
{
int n; // power n
char s[6]; // real number R, at most the length is 6
while (scanf("%s %d", s, &n) != EOF)
{
int position=0, i=0, num=0, j=0;
for (i=0; i<strlen(s); i++)
{
if (s[i] == '.')
{
position = (strlen(s) - 1 - i) * n; // calculate decimal point position after R^n
}
else
{
num = num*10 + s[i] - 48; // transfer float to integer
}
}
// product calculation
product[0]=1;
len = 1;
for (i = 0; i < n; i++)
{
multiply(product, num);
}
// format output
if (len <= position) // product is less than 1
{
printf("0");
printf("."); // print decimal point
for (i=0; i<position-len; i++)
{
printf("0"); // print zero between decimal point and decimal
}
j = 0;
//while (product[j] == 0) // trim trailing zeros
//{
// j++;
//}
for (i=len-1; i>=j; i--)
{
printf("%d", product[i]);
}
}
else
{
j=0;
while (product[j]==0 && j<position) // trim trailing zeros
{
j++;
}
for (i=len-1; i>=j; i--)
{
if (i+1 == position) // cause index in C language starts from 0
{
printf(".");
}
printf("%d", product[i]);
}
}
}
}