1026: [SCOI2009]windy數
Time Limit: 1 Sec Memory Limit: 162 MBSubmit: 4193 Solved: 1877
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Description
windy定義了一種windy數。不含前導零且相鄰兩個數字之差至少爲2的正整數被稱爲windy數。 windy想知道,在A和B之間,包括A和B,總共有多少個windy數?
Input
包含兩個整數,A B。
Output
一個整數。
Sample Input
1 10
【輸入樣例二】
25 50
Sample Output
9
【輸出樣例二】
20
HINT
【數據規模和約定】100%的數據,滿足 1 <= A <= B <= 2000000000 。
裸數位DP,DP[i][j]表示長度爲i以j開頭的windy數有多少個
代碼如下:
/**************************************************************
Problem: 1026
User: duyixian
Language: C++
Result: Accepted
Time:0 ms
Memory:1276 kb
****************************************************************/
/*
* @Author: 逸閒
* @Date: 2015-10-01 16:26:54
* @Last Modified by: 逸閒
* @Last Modified time: 2015-10-01 18:19:12
*/
#include "cstdio"
#include "cstdlib"
#include "iostream"
#include "algorithm"
#include "cstring"
#include "queue"
#include "cmath"
using namespace std;
#define INF 0x3F3F3F3F
#define MAX_SIZE 50
#define Eps
#define Mod
inline int Get_Int()
{
int Num = 0;
char ch;
do
ch = getchar();
while(ch < '0' || ch > '9');
do
{
Num = Num * 10 + ch - '0';
ch = getchar();
}
while(ch >= '0' && ch <= '9');
return Num;
}
int A, B, DP[MAX_SIZE][11];
inline int Count(int Num)
{
int A[MAX_SIZE], Total = 0, Ans = 0;
while(Num)
{
A[++Total] = Num % 10;
Num /= 10;
}
A[Total + 1] = -1;
for(int i = Total; i >= 1; --i)
{
for(int j = 0; j < A[i]; ++j)
if(abs(j - A[i + 1]) > 1)
Ans += DP[i][j];
if(abs(A[i] - A[i + 1]) < 2)
break;
}
for(int i = 1; i < Total; ++i)
for(int j = 1; j <= 9; ++j)
Ans += DP[i][j];
return Ans;
}
int main()
{
for(int i = 0; i <= 9; ++i)
DP[1][i] = 1;
for(int i = 2; i <= 13; ++i)
for(int j = 0; j <= 9; ++j)
{
for(int k = j - 2; k >= 0; --k)
DP[i][j] += DP[i - 1][k];
for(int k = j + 2; k <= 9; ++k)
DP[i][j] += DP[i - 1][k];
}
cin >> A >> B;
cout << Count(B + 1) - Count(A) << endl;
return 0;
}