Bomb HDU - 3555 (数位DP)
Bomb HDU - 3555 (数位DP)
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?
InputThe 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.
OutputFor 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.
题意:求1-n之间有几个数带有49
题解:一题数位DP,可以用记忆化搜索解决,感觉板子还是有点懂了,但不知道改了之后会不会了,具体的dfs的每一步的具体操作已经放在代码里了
#include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<sstream> #include<cmath> #include<stack> #include<cstdlib> #include <vector> #include<queue> using namespace std; using namespace std; #define ms(a,b) memset(a,b,sizeof(a)) #define lson rt*2,l,(l+r)/2 #define rson rt*2+1,(l+r)/2+1,r typedef unsigned long long ull; typedef long long ll; const int MAXN=1e4+5; const double EPS=1e-8; const int INF=0x3f3f3f3f; const int MOD = 1e9+7; #define ll long long #define llu unsigned long long #define INF 0x3f3f3f3f #define PI acos(-1.0) const int maxn = 1e5+5; const int mod = 1e9+7; int bit[40]; ll f[40][4]; ll dp(int pos,int st,bool flag) { //pos为位,st为状态,st=0:没有49,st=1:前一位为4,st=2,表示49都已经出现过了 //flag表示高位与原数是否相同 if(pos == 0) return st == 2; if(flag && f[pos][st] != -1) return f[pos][st]; ll ans = 0; int x; if(flag) x = 9; else x = bit[pos]; //cout<<x<<endl; for(int i = 0;i <= x; i++) { if((st == 2) || (st == 1 && i == 9)) //如果上一位已经有49(st==2)或者上一位为4(st==1)当前位为9 ans += dp(pos-1,2,flag || i<x); //只能加上上一个状态,且上一个状态一定为已经有49了 else if(i == 4) ans += dp(pos-1,1,flag || i<x); //当前为4和前一位也为4的状态是一样的 else //维持st=0的状态 ans += dp(pos-1,0,flag || i<x); } if(flag) f[pos][st] = ans; //记忆化 return ans; } ll calc(ll x) { int len = 0; while(x) { bit[++len] = x % 10; x /= 10; } //cout<<len<<endl; return dp(len,0,0); } int main() { int t; scanf("%d",&t); memset(f,-1,sizeof f); while(t--) { ll n; scanf("%lld",&n); printf("%lld\n",calc(n)); } }