简单的数位dp

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数位dp的模板

hdu 2089

设dp[len][flag]，flag = 1表示前一位是6,否则前一位不是6.

#include <stdio.h>
#include <iostream>
#include <map>
#include <set>
#include <list>
#include <stack>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#include <string>
#include <stdlib.h>
#include <algorithm>
#define LL __int64
#define eps 1e-12
#define PI acos(-1.0)
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 50010;int dig[10];
int dp[10][2];int dfs(int len, int flag, int up)
{if(len == 0)return 1;if(!up && dp[len][flag] != -1)return dp[len][flag];int n = up == 1?dig[len]:9;int res = 0;for(int i = 0; i <= n; i++){if(i == 4 || (flag == 1 && i == 2))continue;res += dfs(len-1,i==6?1:0,up&&(i==n) );}if(!up)dp[len][flag] = res;return res;
}int cal(int num)
{int cnt = 0;while(num){dig[++cnt] = num%10;num /= 10;}memset(dp,-1,sizeof(dp));return dfs(cnt,0,1);
}
int main()
{int a,b;while(~scanf("%d %d",&a,&b)){if(a == 0 && b == 0)break;printf("%d\n",cal(b) - cal(a-1) );}return 0;
}

hdu 3555

不能出现49，与上题类似

#include <stdio.h>
#include <iostream>
#include <map>
#include <set>
#include <list>
#include <stack>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#include <string>
#include <stdlib.h>
#include <algorithm>
#define LL __int64
#define eps 1e-12
#define PI acos(-1.0)
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 4010;int dig[100];
LL dp[100][4];LL dfs(int len, int flag, int up)
{if(len == 0)return flag == 2;if(!up && dp[len][flag] != -1)return dp[len][flag];int n = up ? dig[len] : 9;LL res = 0;for(int i = 0; i <= n; i++){int tflag = flag;if(flag == 0 && i == 4)tflag = 1;else if(flag == 1 && i == 9)tflag = 2;else if(flag == 1 && i != 4)tflag = 0;res += dfs(len-1,tflag,up&&(i==n));}if(!up)dp[len][flag] = res;return res;
}LL cal(LL num)
{int len = 0;while(num){dig[++len] = num % 10;num /= 10;}memset(dp,-1,sizeof(dp));return dfs(len,0,1);
}int main()
{int test;LL n;scanf("%d",&test);while(test--){scanf("%I64d",&n);printf("%I64d\n",cal(n));}return 0;
}

SCOI2009

要求两个相邻的数之差大于等于2，对于第一位是特殊的。所以dfs的时候加一个变量first表示前面是否全是前导0。

#include <stdio.h>
#include <iostream>
#include <map>
#include <set>
#include <list>
#include <stack>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#include <string>
#include <stdlib.h>
#include <algorithm>
#define LL __int64
#define eps 1e-12
#define PI acos(-1.0)
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 4010;int dig[15];
int dp[15][15]; //dp[i][j]表示第i位它的上一位即i+1位为jint dfs(int len, int pre, int up, int first)
{if(len == 0)return first == 0;if(!up && dp[len][pre] != -1 && !first)return dp[len][pre];int n = up ? dig[len] : 9;int res = 0;for(int i = 0; i <= n; i++){if(first) //当前这一位是第一位，当i==0时，第i-1为也是第一位，否则不是res += dfs(len-1,i,up&&i==n,first&&i==0);else if(abs(pre-i) >= 2)res += dfs(len-1,i,up&&i==n,first);}if(!up && !first)dp[len][pre] = res;return res;
}int cal(int num)
{int len = 0;while(num){dig[++len] = num%10;num /= 10;}memset(dp,-1,sizeof(dp));return dfs(len,0,1,1);
}int main()
{int a,b;while(~scanf("%d %d",&a,&b)){printf("%d\n",cal(b) - cal(a-1));}return 0;}

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