题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3555
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 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.
Author
fatboy_cw@WHU
Source
题意:
求0 到n的数中有多少个数字是含有‘49’的。
PS:
数位DP
//dp[i][j]:长度为i的数的第j种状态
//dp[i][0]:长度为i可是不包括49的方案数
//dp[i][1]:长度为i且不含49可是以9开头的数字的方案数
//dp[i][2]:长度为i且包括49的方案数
(转)状态转移例如以下
dp[i][0] = dp[i-1][0] * 10 - dp[i-1][1]; // not include 49 假设不含49且,在前面能够填上0-9 可是要减去dp[i-1][1] 由于4会和9构成49
dp[i][1] = dp[i-1][0]; // not include 49 but starts with 9 这个直接在不含49的数上填个9即可了
dp[i][2] = dp[i-1][2] * 10 + dp[i-1][1]; // include 49 已经含有49的数能够填0-9,或者9开头的填4
接着就是从高位開始统计
在统计到某一位的时候,加上 dp[i-1][2] * digit[i] 是显然对的。由于这一位能够填 0 - (digit[i]-1)
若这一位之前挨着49,那么加上 dp[i-1][0] * digit[i] 也是显然对的。
若这一位之前没有挨着49,可是digit[i]比4大,那么当这一位填4的时候,就得加上dp[i-1][1]
代码例如以下:
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; typedef __int64 LL; LL dp[27][3]; int c[27]; //dp[i][j]:长度为i的数的第j种状态 //dp[i][0]:长度为i可是不包括49的方案数 //dp[i][1]:长度为i且不含49可是以9开头的数字的方案数 //dp[i][2]:长度为i且包括49的方案数 void init() { memset(dp,0,sizeof(dp)); dp[0][0] = 1; for(int i = 1; i <= 20; i++) { dp[i][0] = dp[i-1][0]*10-dp[i-1][1]; dp[i][1] = dp[i-1][0]*1; dp[i][2] = dp[i-1][2]*10+dp[i-1][1]; } } int cal(LL n) { int k = 0; memset(c,0,sizeof(c)); while(n) { c[++k] = n%10; n/=10; } c[k+1] = 0; return k; } void solve(int len, LL n) { int flag = 0;//标记是否出现过49 LL ans = 0; for(int i = len; i >= 1; i--) { ans+=c[i]*dp[i-1][2]; if(flag) { ans+=c[i]*dp[i-1][0]; } else if(c[i] > 4) { //这一位前面没有挨着49。但c[i]比4大,那么当这一位填4的时候,要加上dp[i-1][1] ans+=dp[i-1][1]; } if(c[i+1]==4 && c[i]==9) { flag = 1; } } printf("%I64d\n",ans); } int main() { int t; LL n; init(); scanf("%d",&t); while(t--) { scanf("%I64d",&n); int len = cal(n+1); solve(len, n); } return 0; }
DFS版
代码例如以下:
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define LL __int64 LL n, dp[25][3]; //dp[i][j]:长度为i。状态为j int digit[25]; //nstatus: 0:不含49, 1:不含49但末尾是4, 2 :含49 LL DFS(int pos, int status, int limit) { if(pos <= 0) // 假设到了已经枚举了最后一位。而且在枚举的过程中有49序列出现 return status==2;//注意是 == if(!limit && dp[pos][status]!=-1) //对于有限制的询问我们是不可以记忆化的 return dp[pos][status]; LL ans = 0; int End = limit?digit[pos]:9; // 确定这一位的上限是多少 for(int i = 0; i <= End; i++) // 每一位有这么多的选择 { int nstatus = status; // 有点else s = statu 的意思 if(status==0 && i==4)//高位不含49。而且末尾不是4 ,如今末尾添4返回1状态 nstatus = 1; else if(status==1 && i!=4 && i!=9)//高位不含49。且末尾是4,如今末尾加入的不是4返回0状态 nstatus = 0; else if(status==1 && i==9)//高位不含49,且末尾是4,如今末尾加入9返回2状态 nstatus = 2; ans+=DFS(pos-1, nstatus, limit && i==End); } if(!limit) dp[pos][status]=ans; return ans; } int cal(LL x) { int cnt = 0; while(x) { digit[++cnt] = x%10; x/=10; } digit[cnt+1] = 0; return cnt; } int main() { int t; scanf("%d",&t); while(t--) { memset(dp,-1,sizeof(dp)); scanf("%I64d",&n); int len = cal(n); LL ans = DFS(len, 0, 1); printf("%I64d\n",ans); } return 0; }