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3 1 50 500
0 1 15HintFrom 1 to 500, the numbers that include the sub-sequence "49" are "49","149","249","349","449","490","491","492","493","494","495","496","497","498","499", so the answer is 15.
#include<stdio.h> #include<string.h> int bit[20]; __int64 dp[20][10][2]; __int64 dfs(int pos,int pre,int isture,int limit) { if(pos<0) { return isture; } if(!limit&&dp[pos][pre][isture]!=-1) { return dp[pos][pre][isture]; } int last=limit?bit[pos]:9; __int64 ans=0; for(int i=0;i<=last;i++) { ans+=dfs(pos-1,i,isture||(pre==4&&i==9),limit&&(i==last)); } if(!limit) { dp[pos][pre][isture]=ans; } return ans; } __int64 solve(__int64 n) { int len=0; while(n) { bit[len++]=n%10; n/=10; } return dfs(len-1,0,0,1); } int main() { int n; memset(dp,-1,sizeof(dp)); int t; scanf("%d",&t); while(t--) { __int64 n; scanf("%I64d",&n); printf("%I64d\n",solve(n)); } }人家的代码http://blog.csdn.net/scf0920/article/details/42870573
#include <iostream> #include <string.h> #include <math.h> #include <queue> #include <algorithm> #include <stdlib.h> #include <map> #include <set> #include <stdio.h> using namespace std; #define LL __int64 #define pi acos(-1.0) const int mod=100000000; const int INF=0x3f3f3f3f; const double eqs=1e-8; LL dp[21][11], c[21]; LL dfs(int cnt, int pre, int maxd, int zero) { if(cnt==-1) return 1; if(maxd&&zero&&dp[cnt][pre]!=-1) return dp[cnt][pre]; int i, r; LL ans=0; r=maxd==0?c[cnt]:9; for(i=0;i<=r;i++){ if(!zero||!(pre==4&&i==9)){ ans+=dfs(cnt-1,i,maxd||i<r,zero||i); } } if(maxd&&zero) dp[cnt][pre]=ans; return ans; } LL Cal(LL x) { int i, cnt=0; while(x){ c[cnt++]=x%10; x/=10; } return dfs(cnt-1,-1,0,0); } int main() { int t; LL n; scanf("%d",&t); memset(dp,-1,sizeof(dp)); while(t--){ scanf("%I64d",&n); printf("%I64d\n",n+1-Cal(n)); } return 0; }
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原文地址:http://blog.csdn.net/yu_ch_sh/article/details/47130147