UESTC 250 windy数(数位DP)
2015-07-26 17:35
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题意:题意:求区间[A,B]之间的,不含前导0,且相邻两数位之间相差至少为2的正整数有多少个.
分析:dp[i][j]表示,长度为i、以j为结尾的表示的个数,再加一个前导0判断即可
分析:dp[i][j]表示,长度为i、以j为结尾的表示的个数,再加一个前导0判断即可
#include <map> #include <set> #include <list> #include <cmath> #include <queue> #include <stack> #include <cstdio> #include <vector> #include <string> #include <cctype> #include <complex> #include <cassert> #include <utility> #include <cstring> #include <cstdlib> #include <iostream> #include <algorithm> using namespace std; typedef pair<int,int> PII; typedef long long ll; #define lson l,m,rt<<1 #define pi acos(-1.0) #define rson m+1,r,rt<<11 #define All 1,N,1 #define read freopen("in.txt", "r", stdin) const ll INFll = 0x3f3f3f3f3f3f3f3fLL; const int INF= 0x7ffffff; const int mod = 1000000007; int dp[15][15],a,b; int bit[15],len; //f表示前导0是否有效 int dfs(int i,int j,int f,int e){ if(i==0) { if(f)return 0; else return 1; } if(!f&&!e&&dp[i][j]!=-1)return dp[i][j]; int l=e?bit[i]:9; int num=0; for(int v=0;v<=l;++v){ if(f){ if(v==0) num+=dfs(i-1,v,1,e&&(v==l)); else num+=dfs(i-1,v,0,e&&(v==l)); } else if(abs(j-v)>=2) num+=dfs(i-1,v,0,e&&(v==l)); } if(!f&&!e)dp[i][j]=num; return num; } int solve(ll x){ len=0; while(x){ bit[++len]=x%10; x/=10; } return dfs(len,0,1,1); } int main() { memset(dp,-1,sizeof(dp)); scanf("%d%d",&a,&b); printf("%d\n",solve(b)-solve(a-1)); return 0; }
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