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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;
} 版权声明:本文为博主原创文章,未经博主允许不得转载。
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原文地址:http://blog.csdn.net/yu_ch_sh/article/details/47130147