HDU-3555 Bomb （数位DP）

Bomb

http://acm.hdu.edu.cn/showproblem.php?pid=3555

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)



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.

题目大意：统计1~n中含有连续49的数字的个数？

数位DP入门题

设dp[i][0]表示长度为i，不含49的数字个数；dp[i][1]表示长度为i，不含49且最高位为9的数字个数；dp[i][2]表示长度为i，含有49的数字个数

不是很明白为什么第二维状态要这样取...

#include <cstdio>

using namespace std;

int num[25],len;
long long n,ans;
long long dp[25][3];//dp[i][0]表示长度为i，不含49的数字个数；dp[i][1]表示长度为i，不含49且最高位为9的数字个数；dp[i][2]表示长度为i，含有49的数字个数

void Init() {
dp[0][0]=1;
dp[0][1]=dp[0][2]=0;
for(int i=1;i<=20;++i) {
dp[i][0]=dp[i-1][0]*10-dp[i-1][1];//长度为i-1且不含49的，最高位加上0~9，共10种情况；但若最高位加上4，则要减去次高位为9的情况dp[i-1][1]
dp[i][1]=dp[i-1][0];//长度i-1且不含49的，最高位加上9
dp[i][2]=dp[i-1][2]*10+dp[i-1][1];//长度i-1且含49的，最高位加上0~9；长度i-1且最高位为9的，最高位加上4
}
}

long long getCnt(long long x) {//得到区间[1,x]中满足题意的数字个数
++x;
ans=len=0;
while(x>0) {
num[++len]=x%10;
x/=10;
}
num[len+1]=-1;
bool flag=false;//判断高位是否出现过49，true表示高位出现过
for(int i=len;i>=1;--i) {//从最高位开始枚举
ans+=dp[i-1][2]*num[i];//当前位取0~(num[i]-1)，后面i-1位数为含49的数

if(flag) {//如果高位已经出现过49
ans+=dp[i-1][0]*num[i];//当前位去0~(num[i]-1)，后面i-1位数为不含49的数【最开始已经加上了含49的数，所以不必再加】
}
if(!flag&&num[i]>4) {//如果高位未出现过49，且当前位为大于4
ans+=dp[i-1][1];
}
if(num[i+1]==4&&num[i]==9) {//若该位与上一位能组成49
flag=true;
}
}
return ans;
}

int main() {
Init();
int T;
scanf("%d",&T);
while(T-->0) {
scanf("%I64d",&n);
printf("%I64d\n",getCnt(n));
}
return 0;
}