Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 11804 Accepted Submission(s): 4212
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?
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.
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
1
50
500
Sample Output
0
1
15
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.
Author
fatboy_cw@WHU
Source
【思路】
数位DP。
预处理：
f[i][0] 表示i位数中不含49的数的数目
f[i][1] 表示i位数中不含49且最高位为9的数的数目
f[i][2] 表示i位数中含49的数的数目
根据n的每一位统计ans，具体见代码。
【代码】
#include<cstdio>
using namespace std; typedef long long LL;
LL f[][];
/*
f[i][0] i位数 无49存在
f[i][1] i位数 无49存在且末尾为9
f[i][2] i位数 有49
*/
int b[]; void init() {
f[][]=; f[][]=f[][]=;
for(int i=;i<;i++) {
f[i][]=*f[i-][]-f[i-][];
f[i][]=f[i-][];
f[i][]=*f[i-][]+f[i-][];
}
} int main() {
int T; LL n;
scanf("%d",&T);
init();
while(T--) {
scanf("%I64d",&n);
int len=;
while(n)
b[++len]=n% , n/=;
b[len+]=;
LL ans=; bool flag=;
for(int i=len;i;i--) {
ans += b[i]*f[i-][]; // + ...49...
if(flag) ans += f[i-][]*b[i]; // + 49...(无49...)
else if(b[i]>) ans += f[i-][]; // + 4(9 无49)
if(b[i+]== && b[i]==) flag=;
}
if(flag) ans++; //算上自身
printf("%I64d\n",ans);
}
return ;
}