「hdu3555」Bomb
Problem Description
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?
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.
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 50 500
Sample 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.
题解
复习数位dp。。。
好像我的写法和网上都不太一样
预处理f[i][j]表示长度为i开头为j的不满足条件的数字个数
ans=n-dp(n)
然后就可以dp了。。。诶好像是废话
顺便说下dp过程
比如78915 L=5
首先算出 1-9999
for(int i=1;i<L;i++)
for(int j=1;j<=9;j++)
ans+=f[i][j];
for(int j=1;j<=9;j++)
ans+=f[i][j];
然后是10000-69999
for(int i=1;i<cur;i++)ans+=f[L][i];
现在限定第一位为7，算70000-77999
。。。。第二位位8，算78000-78899
以此类推
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#include<iostream> #include<cstdio> #include<cstdlib> #include<algorithm> #include<cstring> #include<cmath> #define inf 0x7fffffff #define ll long long using namespace std; inline ll read() { ll x=0;char ch=getchar(); while(ch>'9'||ch<'0')ch=getchar(); while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x; } ll base[20]; int T; ll n,f[20][10]; void pre() { base[1]=1;for(int i=2;i<=19;i++)base[i]=base[i-1]*10; for(int i=0;i<=9;i++)f[1][i]=1; for(int i=2;i<=19;i++) for(int j=0;j<=9;j++) for(int k=0;k<=9;k++) if(j!=4||k!=9) f[i][j]+=f[i-1][k]; } ll dp(ll x) { int L=19,cur,pre; ll ans=0; while(base[L]>x)L--; for(int i=1;i<L;i++) for(int j=1;j<=9;j++) ans+=f[i][j]; cur=x/base[L];x%=base[L]; for(int i=1;i<cur;i++)ans+=f[L][i]; if(L==1)ans++; pre=cur; for(int i=L-1;i;i--) { cur=x/base[i];x%=base[i]; if(i==1) { for(int j=0;j<=cur;j++) if(pre!=4||j!=9)ans+=f[i][j]; } else { for(int j=0;j<cur;j++) if(pre!=4||j!=9)ans+=f[i][j]; } if(pre==4&&cur==9)break; pre=cur; } return ans; } int main() { pre(); T=read(); while(T--) { n=read(); cout<<n-dp(n)<<endl; } return 0; } |
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计数器 9354907
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