高精度大数据计算,这道题很拟真,数据问题做了半天。
//
// main.cpp
// ACMUsing
//
// Created by Nsamper on 2017/1/4.
// Copyright © 2017年 Nsamper. All rights reserved.
//
#include
#include
#include
#include
#include
//1001
using namespace std;
string r;
string ans;
int n;
long pointNum;
string eraseZero(string str) {
long i;
pointNum = 0;
long point = str.length();
while (str[0] == '0') {
str.erase(str.begin());
}
for (i = 0; i < str.length(); i++) {
if (str[i] == '.') {
point = i;//小数点的位置
break;
}
}
if (point == str.length()) {
return str;
}
for (i = str.length()-1; i > point; i--) {
if (str[i] != '0') {
break;
} else {
str.erase(str.begin()+i);
point--;
}
}
return str;
}
void removePoint() {
int i = 0;
for (i = 0; i < r.length(); i++) {
if (r[i] == '.') {
pointNum = r.length() - i - 1;//小数的个数
r.erase(r.begin()+i);
}
}
}
string multify(string a, string b) {
int i,j;
int lena,lenb;
int tmp[1000];
string tempStr;
int x,y = 0;
reverse(a.begin(), a.end());
reverse(b.begin(), b.end());
memset(tmp, 0, sizeof(tmp));
lena = (int)a.length();
lenb = (int)b.length();
for (i = 0; i < lena; i ++) {
for (j = 0; j < lenb; j ++) {
x = a[i]-'0';
y = b[j]-'0';
tmp[i + j] += x * y;
}
}
for (i = 0; i < lena + lenb - 1; i ++) {
x = tmp[i] % 10;
y = tmp[i] / 10;
tmp[i] = x;
tmp[i + 1] += y;
}
if (y > 0) {
i++;
}
for (j = 0; j < i; j ++) {
tempStr += tmp[j]+'0';
}
reverse(tempStr.begin(), tempStr.end());
return tempStr;
}
void work(string a, int len) {
ans = "1";
while (len) {
if (len&1) {
ans = multify(ans, a);
}
a = multify(a, a);
len >>= 1;
}
}
void eraseZeroAndPrint(string str) {
str = eraseZero(str);
if (str[str.length()-1] == '.') {
str.erase(str.length()-1);
}
cout<> r >> n) {
str = "";
r = eraseZero(r);
removePoint();
for(i=0;i ans.length()) {
i = pointNum*n - (int)ans.length();
str+='.';
while (i > 0) {
str+='0';
i--;
}
for (j = 0; j < ans.length(); j ++) {
str+= ans[j];
}
} else {
for (j = 0; j < ans.length(); j ++) {
if (j == ans.length() - pointNum*n) {
str+='.';
}
str+=ans[j];
}
}
eraseZeroAndPrint(str);
}
return 0;
}