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保存代码In Reply To:枚举前面若干位为首个数字n的结尾,那么后续的数字肯定是n+1,n+2,n+3...一整个组成的,枚举这个数字即可 Posted by:yygy at 2023-05-11 17:05:17 // MyFirstApp.cpp : 此文件包含 "main" 函数。程序执行将在此处开始并结束。
//
#include <iostream>
#include <vector>
#include <math.h>
#include <algorithm>
using namespace std;
typedef __int64 lld;
const int MAXN = 20010;
struct BigNum
{
int dig[512];//题目说最多200位,平方之后最多400位
int len;
BigNum() {}
BigNum(int arr[], int size)
{
clr();
len = size;
for (int i = 0; i < len; i++)
{
dig[len - i - 1] = arr[i];
}
}
void clr()
{
memset(dig, 0, sizeof(int) * 512);
len = 1;
}
void operator --()
{
dig[0]--;
int bit=0;
while (dig[bit] < 0)
{
dig[bit] += 10;
dig[bit + 1] --;
}
while (len > 1 && dig[len - 1] == 0)
{
len--;
}
}
bool operator<(const BigNum& other) const
{
if (len < other.len)
{
return true;
}
if (other.len < len)
{
return false;
}
for (int i = 0; i < len; i++)
{
if (dig[i] != other.dig[i])
{
return dig[i] < other.dig[i];
}
}
return false;
}
void operator ++()
{
dig[0]++;
int bit = 0;
while (dig[bit] > 9)
{
dig[bit] -= 10;
dig[bit + 1] ++;
}
if (dig[len])
{
len++;
}
}
};
bool IsSuffix(const BigNum& num, int* pat, int size)
{
if (size > num.len)
{
return false;
}
for (int i = 0; i < size; i++)
{
if (num.dig[i] != pat[size - i - 1])
{
return false;
}
}
return true;
}
bool Match(int* pat, int patlen, const BigNum& tmp, int last)
{
int i = 0;
for (i = 0; i < tmp.len && last + i < patlen; i++)
{
if (tmp.dig[i] != pat[last + i])
{
return false;
}
}
return i == tmp.len || last + i == patlen;
}
void FindSeperateSeveralSegments(int patlen, int* pat, BigNum& ans, int& offset)
{
//把pat分解成前后两部分,后面一部分是由n,n+1,n+2...堆起来的,那么后面的开头数字长度不能小于第一部分
for (int i = 0; patlen - (i + 1) >= i + 1; i++)
{
if (pat[i + 1] == 0)
{
//首位数字必须非0
continue;
}
//枚举后面数字[i+1,j]区间为第一个数字
//j-i>=i+1
for (int j = i + i + 1; j < patlen; j++)
{
BigNum head = BigNum(pat + i + 1, j - i);
BigNum pre = head;
--pre;
//判断前面部分是否为head-1的后缀
if (!IsSuffix(pre, pat, i + 1))
{
continue;
}
int last = j + 1;
BigNum tmp = head;
//tmp一直++,把last后面的数字依依匹配,看看能不能配完
while (last < patlen)
{
++tmp;
if (!Match(pat, patlen, tmp, last))
{
break;
}
last += tmp.len;
}
if (last >= patlen)//匹配成功
{
//记录一下现在的答案
if (pre < ans)
{
ans = pre;
int newOffset = 1 + pre.len - (i + 1);
offset = newOffset;
}
}
}
}
}
int main()
{
char str[512];
int pat[512];
int patlen = 0;
while (cin.getline(str, 1024))
{
patlen = strlen(str);
for (int i = 0; i < patlen; i++)
{
pat[i] = str[i] - '0';
}
//一个初始解,也就是仅由于段组成的
BigNum ans = BigNum(pat, patlen);
int offset = 1;
if (ans.dig[ans.len - 1] == 0)
{
//最高位为0,需要添加一位,比如0,是10后面来的,01应该是101后面来的
offset++;
ans.dig[ans.len] = 1;
ans.len++;
}
//假设由若干个数字拼成,寻找答案
FindSeperateSeveralSegments(patlen, pat, ans, offset);
//假设只由两个数字拼成,寻找答案
//FindSeperateTwoSegments(patlen, pat, ans, offset);
}
return 0;
}
/*
http://poj.org/problem?id=1863
Subnumber
要考虑前导0,一堆0的情况
*/
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