HDU3555:Bomb(数位DP)
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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
3150500
Sample Output
0115HintFrom 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:与HDU2089一样的意思,要求1~n的范围内含有49的数字的个数
思路:按照2089的思路去做就好了
#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;__int64 dp[25][3];void Init(){ memset(dp,0,sizeof(dp)); dp[0][0] = 1; int i; for(i = 1;i<=22;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]; }}__int64 solve(__int64 n){ __int64 i,tem = n,len = 0,a[25],flag = 0,ans = 0; while(n) { a[++len] = n%10; n/=10; } a[len+1] = 0; for(i = len;i;i--) { ans+=dp[i-1][2]*a[i]; if(flag) ans+=dp[i-1][0]*a[i]; if(!flag && a[i]>4) ans+=dp[i-1][1]; if(a[i+1] == 4 && a[i] == 9) flag = 1; } return ans;}int main(){ int t; __int64 n; scanf("%d",&t); Init(); while(t--) { scanf("%I64d",&n); printf("%I64d\n",solve(n+1)); } return 0;}
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