不要62
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 56059 Accepted Submission(s): 21685
Problem Description
杭州人称那些傻乎乎粘嗒嗒的人为62（音：laoer）。
杭州交通管理局经常会扩充一些的士车牌照，新近出来一个好消息，以后上牌照，不再含有不吉利的数字了，这样一来，就可以消除个别的士司机和乘客的心理障碍，更安全地服务大众。
不吉利的数字为所有含有4或62的号码。例如：
62315 73418 88914
都属于不吉利号码。但是，61152虽然含有6和2，但不是62连号，所以不属于不吉利数字之列。
你的任务是，对于每次给出的一个牌照区间号，推断出交管局今次又要实际上给多少辆新的士车上牌照了。
Input
输入的都是整数对n、m（0<n≤m<1000000），如果遇到都是0的整数对，则输入结束。
Output
对于每个整数对，输出一个不含有不吉利数字的统计个数，该数值占一行位置。
Sample Input
1 100
0 0Sample Output
80#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<iomanip>
#include<algorithm>
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
#define swap(a,b) (a=a+b,b=a-b,a=a-b)
#define memset(a,v) memset(a,v,sizeof(a))
#define X (sqrt(5)+1)/2.0 //Wythoff
#define Pi acos(-1)
#define eps 1.0e-8
using namespace std;
typedef long long ll;
const int MAXL(1e5);
const int INF(0x3f3f3f3f);
ll dp[10][20];
void getdp()
{
memset(dp,0);
int i,j,k;
dp[0][0]=1;
for(i=1;i<=7;i++)
{
for(j=0;j<=9;j++)
{
if(j==4)
dp[i][j]=0;
else
if(j==6)
{
for(k=0;k<=9;k++)
{
if(k!=2)
dp[i][j]+=dp[i-1][k];
}
}
else
{
for(k=0;k<=9;k++)
dp[i][j]+=dp[i-1][k];
}
}
}
}
ll a[20];
ll solve(ll n)
{
ll i,j,e=1;
while(n)
{
a[e++]=n%10;
n/=10;
}
a[e]=0;
ll ans=0;
for(i=e-1;i>=1;i--)
{
for(j=0;j<a[i];j++)
{
if(j==4||a[i+1]==6&&j==2)
continue;
else
ans+=dp[i][j];
}
if(a[i]==4)
break;
if(a[i+1]==6&&a[i]==2)
break;
}
return ans;
}
int main()
{
ll i,j,ans1,ans2,n,m;
getdp();
while(~scanf("%lld%lld",&n,&m)&&(n||m))
{
ans1=solve(n);
ans2=solve(m+1);
printf("%lld
",ans2-ans1);
}
return 0;
}
Bomb
The counter-terrorists found a time bomb in the dust. But this time the terrorists improve on the time bomb. The number sequence of the time bomb counts from 1 to N. If the current number sequence includes the sub-sequence "49", the power of the blast would add one point.
Now the counter-terrorist knows the number N. They want to know the final points of the power. Can you help them?
Input
The first line of input consists of an integer T (1 <= T <= 10000), indicating the number of test cases. For each test case, there will be an integer N (1 <= N <= 2^63-1) as the description.
The input terminates by end of file marker.
Output
For each test case, output an integer indicating the final points of the power.
Sample Input
3
1
50
500Sample Output
0
1
15 Hint
From 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.
2的63次方是9223372036854775808（19位）
思路：可以用上面的模板将49相邻的数计算出来 减掉就行
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<iomanip>
#include<algorithm>
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
#define swap(a,b) (a=a+b,b=a-b,a=a-b)
#define memset(a,v) memset(a,v,sizeof(a))
#define X (sqrt(5)+1)/2.0 //Wythoff
#define Pi acos(-1)
#define eps 1.0e-8
using namespace std;
typedef long long ll;
const int MAXL(1e5);
const int INF(0x3f3f3f3f);
ll dp[20][20];
void getdp()
{
memset(dp,0);
int i,j,k;
dp[0][0]=1;
for(i=1;i<=20;i++)
{
for(j=0;j<=9;j++)
{
if(j==4)
{
for(k=0;k<=9;k++)
{
if(k!=9)
dp[i][j]+=dp[i-1][k];
}
}
else
{
for(k=0;k<=9;k++)
dp[i][j]+=dp[i-1][k];
}
}
}
}
ll a[20];
ll solve(ll n)
{
ll i,j,e=1;
while(n)
{
a[e++]=n%10;
n/=10;
}
a[e]=0;
ll ans=0;
for(i=e-1;i>=1;i--)
{
for(j=0;j<a[i];j++)
{
if(a[i+1]==4&&j==9)
continue;
else
ans+=dp[i][j];
}
if(a[i+1]==4&&a[i]==9)
break;
}
return ans;
}
int main()
{
ll i,j,ans1,ans2,n,m,T;
getdp();
scanf("%lld",&T);
while(T--)
{
scanf("%lld",&n);
ans1=solve(n+1);
printf("%lld
",n-ans1+1);
}
return 0;
}
//可以改成这样
//ans1=solve(n+1)-solve(1);
//printf("%lld
",n-ans);