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hdu 3555 Bomb ( 数位DP)
2018-03-21 03:57
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Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 7926 Accepted Submission(s): 2780
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?
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.
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
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.
思路:
dp[i][0]表示不含49的个数 。
dp[i][1]表示不含49且高位为9的个数;
dp[i][2]表示包括49的个数;
dp[i][0]=10*dp[i-1][0]-dp[i-1][1]; //不含49的数能够随意加10个数字。减去9前面加4的个数
dp[i][1]=dp[i-1][0]; //不含49的数最高位加9
dp[i][2]=dp[i-1][2]*10+dp[i-1][i]; //包括49的数字能够加入0~9,高位为9的能够加4;
#include"stdio.h" #include"string.h" #include"iostream" #include"algorithm" #include"math.h" #include"vector" using namespace std; #define LL __int64 #define N 25 LL dp [3]; void inti() { memset(dp,0,sizeof(dp)); dp[0][0]=1; int i; for(i=1; i<N; 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]; } } LL work(LL x) { int i,k,flag; int a ; LL ans; k=0; while(x) { a[++k]=x%10; x/=10; } a[k+1]=ans=flag=0; for(i=k; i>0; i--) { ans+=a[i]*dp[i-1][2]; if(flag) ans+=a[i]*dp[i-1][0]; else { if(a[i]>4) ans+=dp[i-1][1]; } if(a[i+1]==4&&a[i]==9) flag=1; // printf("%d\n",ans); } return ans; } int main() { int T; LL n; inti(); scanf("%d",&T); while(T--) { scanf("%I64d",&n); printf("%I64d\n",work(n+1)); } return 0; }
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