1 高精度(含大数开方)+二分
一个技巧和三点注意:
技巧:假设k^n=p;(k的n次方),那么p的位数/n得到的是k的位数!例如:n=7,p=4357186184021382204544,p的位数为22,用22/7的结果向上取整,得到4,即为k的位数,也就是说k的取值范围是1000~9999。(引自code_pang)不利用这一点,高精度+直接二分,也会超时。用这一个技巧合理缩小二分的范围。
注意:看code的main中的注释。
(二分思想不熟练,因为二分算法很高效,所以一定要暴力点直接确定left和right,然后根据情况,优化也是将left变大和right变小,而不会是其他奇怪的情况。另外,注意left=mid+1,right=mid-1,这很重要)
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
const int numlen = 105; // 位数
int max(int a, int b) { return a>b?a:b; }
struct bign {
int len, s[numlen];
bign() {
memset(s, 0, sizeof(s));
len = 1;
}
bign(int num) { *this = num; }
bign(const char *num) { *this = num; }
bign operator = (const int num) {
char s[numlen];
sprintf(s, "%d", num);
*this = s;
return *this;
}
bign operator = (const char *num) {
len = strlen(num);
while(len > 1 && num[0] == '0') num++, len--;
for(int i = 0;i < len; i++) s[i] = num[len-i-1] - '0';
return *this;
}
void deal() {
while(len > 1 && !s[len-1]) len--;
}
bign operator + (const bign &a) const {
bign ret;
ret.len = 0;
int top = max(len, a.len) , add = 0;
for(int i = 0;add || i < top; i++) {
int now = add;
if(i < len) now += s[i];
if(i < a.len) now += a.s[i];
ret.s[ret.len++] = now%10;
add = now/10;
}
return ret;
}
bign operator - (const bign &a) const {
bign ret;
ret.len = 0;
int cal = 0;
for(int i = 0;i < len; i++) {
int now = s[i] - cal;
if(i < a.len) now -= a.s[i];
if(now >= 0) cal = 0;
else {
cal = 1; now += 10;
}
ret.s[ret.len++] = now;
}
ret.deal();
return ret;
}
bign operator * (const bign &a) const {
bign ret;
ret.len = len + a.len;
for(int i = 0;i < len; i++) {
for(int j = 0;j < a.len; j++)
ret.s[i+j] += s[i]*a.s[j];
}
for(int i = 0;i < ret.len; i++) {
ret.s[i+1] += ret.s[i]/10;
ret.s[i] %= 10;
}
ret.deal();
return ret;
}
//乘以小数,直接乘快点
bign operator * (const int num) {
bign ret;
ret.len = 0;
int bb = 0;
for(int i = 0;i < len; i++) {
int now = bb + s[i]*num;
ret.s[ret.len++] = now%10;
bb = now/10;
}
while(bb) {
ret.s[ret.len++] = bb % 10;
bb /= 10;
}
ret.deal();
return ret;
}
bign operator / (const bign &a) const {
bign ret, cur = 0;
ret.len = len;
for(int i = len-1;i >= 0; i--) {
cur = cur*10;
cur.s[0] = s[i];
while(cur >= a) {
cur -= a;
ret.s[i]++;
}
}
ret.deal();
return ret;
}
bign operator % (const bign &a) const {
bign b = *this / a;
return *this - b*a;
}
bign operator += (const bign &a) { *this = *this + a; return *this; }
bign operator -= (const bign &a) { *this = *this - a; return *this; }
bign operator *= (const bign &a) { *this = *this * a; return *this; }
bign operator /= (const bign &a) { *this = *this / a; return *this; }
bign operator %= (const bign &a) { *this = *this % a; return *this; }
bool operator < (const bign &a) const {
if(len != a.len) return len < a.len;
for(int i = len-1;i >= 0; i--) if(s[i] != a.s[i])
return s[i] < a.s[i];
retu