Bomb(HDU 3555 数位DP)
2017-10-15 20:34
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Bomb
[align=center]Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 20230 Accepted Submission(s): 7565
[/align]
[align=left]Problem Description[/align]
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?
[align=left]Input[/align]
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.
[align=left]Output[/align]
For each test case, output an integer indicating the final points of the power.
[align=left]Sample Input[/align]
3
1
50
500
[align=left]Sample Output[/align]
0
1
15
HintFrom 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:给一个数字n,问1-n里有几个数包含49.
//思路:数位DP:
dp[i][0]代表长度为 i 并且不含有49的数字的个数;
dp[i][1]代表长度为 i 并且不含有49,但是最高位是9的数字的个数;
dp[i][2]代表长度为 i 并且含有49的数字的个数。
状态转移方程:
dp[i][0] = dp[i-1][0] * bit[i] - dp[i-1][1];
dp[i][1]=dp[i-1][1];
dp[i][2]=dp[i-1][2]*bit[i]+dp[i-1][1];
#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> using namespace std; typedef long long ll; ll n; int bit[25]; ll dp[25][5]; int main() { int T; scanf("%d",&T); memset(dp,0,sizeof(dp)); dp[0][0]=1; for(int i=1;i<21;i++) { dp[i][0]=dp[i-1][0]*10-dp[i-1][1]; dp[i][1]=dp[i-1][0]; dp[i][2]=dp[i-1][2]*10+dp[i-1][1]; } while(T--) { scanf("%I64d",&n); memset(bit,0,sizeof(bit)); ll x=n+1; int len=0; while(x) { bit[++len]=x%10; x=x/10; } ll ans=0; bool flag=false; int last=0; for(int i=len;i>=1;i--) { ans+=(dp[i-1][2]*bit[i]); if(flag) ans+=dp[i-1][0]*bit[i]; if(!flag&&bit[i]>4) ans+=dp[i-1][1]; if(last==4&&bit[i]==9) flag=true; last=bit[i]; } printf("%I64d\n",ans); } return 0; }
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