【HDU5898 2016 ACM ICPC Asia Regional Shenyang Online G】【数位DP】odd-even number 范围内有多少数字满足奇串长为偶数偶串长为奇数.

最新推荐文章于 2018-11-10 11:22:00 发布

snowy_smile

最新推荐文章于 2018-11-10 11:22:00 发布

阅读量612

点赞数

分类专栏：题库-HDU 动态规划-数位DP 文章标签：题库-HDU 动态规划-数位DP

本文链接：https://blog.youkuaiyun.com/snowy_smile/article/details/52619409

版权

题库-HDU 同时被 2 个专栏收录

144 篇文章

订阅专栏

动态规划-数位DP

3 篇文章

订阅专栏

本文介绍了一种使用数位动态规划方法解决特定数学问题的算法。该算法用于计算一个区间内符合特定条件（即连续的偶数段长度为奇数，连续的奇数段长度为偶数）的整数数量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

odd-even number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 594 Accepted Submission(s): 327

Problem Description

For a number,if the length of continuous odd digits is even and the length of continuous even digits is odd,we call it odd-even number.Now we want to know the amount of odd-even number between L,R(1<=L<=R<= 9*10^18).

Input

First line a t,then t cases.every line contains two integers L and R.

Output

Print the output for each case on one line in the format as shown below.

Sample Input

Sample Output

  
  
   
   Case #1: 29
Case #2: 36

Source

2016 ACM/ICPC Asia Regional Shenyang Online

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
const int N = 0, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int a[20];
//oddnum要为0，evennum要为1
LL f[64][2][3];
//oddnum和evennum为之前传过来的段数
LL dp(int p, bool tp, int len, int last)
{
	if (p == -1)return len ^ last;
	if (!tp && ~f[p][len][last])return f[p][len][last];
	LL ret = 0;
	int top = tp ? a[p] : 9;
	for (int i = 0; i <= top; ++i)
	{
		if (i == 0 && last == 2)
		{
			ret += dp(p - 1, tp&i == top, 0, 2);
			continue;
		}
		int x = i & 1;
		if (x == last)ret += dp(p - 1, tp&i == top, len ^ 1, x);
		else if (len ^ last)ret += dp(p - 1, tp&i == top, 1, x);
	}
	if (!tp)f[p][len][last] = ret;
	return ret;
}
LL DP(LL x)
{
	int p = -1; while (x)a[++p] = x % 10, x /= 10;
	if (p == -1)a[++p] = 0;
	MS(f, -1);
	return dp(p, 1, 0, 2);
}
int main()
{
	scanf("%d", &casenum);
	for (casei = 1; casei <= casenum; ++casei)
	{
		LL l, r;
		scanf("%lld%lld", &l, &r);
		LL ans = DP(r) - DP(l - 1);
		printf("Case #%d: %lld\n", casei, ans);
	}
	return 0;
}
/*
【题意】
求区间范围内有多少个数，使得该数连续的偶数段的长度都为奇数，连续的奇数段的长度都为偶数

【类型】
数位DP

【分析】
这道题是数位DP
数位DP常常是把高位的状态向低位传。
同时常常要考虑前导零


*/