hdu 3555 Bomb【数位dp~吖!!!!!】
2016-04-12 13:03
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Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 12675 Accepted Submission(s): 4525
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
Author
fatboy_cw@WHU
题目大意:
给你一个数据n,让你求出(0-n]之间包含49的数据个数。
思路:求不包含49的数据个数,然后再用n-这个个数即可,处理方法:数位dp、
我们设dp【i】【j】表示长度为i的,第i位上数据为j的个数,然后一点一点一点一点一点一点慢慢加起来。
预处理dp【i】【j】:
void init() { memset(dp,0,sizeof(dp)); dp[0][0]=1; for(int i=1;i<=21;i++)//枚举长度 { for(int j=0;j<10;j++)//枚举第i位上的数据 { for(int k=0;k<10;k++)//枚举第i-1位上的数据。 { if(!(j==4&&k==9))//合法的情况下累加 { dp[i][j]+=dp[i-1][k];//长度为i,数据为j的情况数。 } } } } }AC代码:
这里因为solve()函数求的是(0-n)的数据,所以在处理的时候n要+1,因为long long 的数据是到2^63-1的,所以我们避免2^63会出现不对的数据,我们开llu。
#include<stdio.h> #include<string.h> using namespace std; #define ll unsigned long long ll dp[25][10]; ll digit[25]; void init() { memset(dp,0,sizeof(dp)); dp[0][0]=1; for(int i=1;i<=21;i++) { for(int j=0;j<10;j++) { for(int k=0;k<10;k++) { if(!(j==4&&k==9)) { dp[i][j]+=dp[i-1][k];//长度为i,最后数字为j的情况数。 } } } } } ll calchangdu(ll n) { ll cont=0; while(n) { cont++; n/=10; } return cont; } ll caldigit(ll n,ll len) { memset(digit,0,sizeof(digit)); for(int i=1;i<=len;i++) { digit[i]=n%10; n/=10; } } ll solve(ll n) { ll ans=0; ll len=calchangdu(n);//计算出n的长度 caldigit(n,len);//计算出n的每一位数据 for(int i=len;i>=1;i--) { for(int j=0;j<digit[i];j++) { if(!(digit[i+1]==4&&j==9)) { ans+=dp[i][j]; } } if(digit[i]==9&&digit[i+1]==4)break; } return ans; } int main() { init(); int t; scanf("%d",&t); while(t--) { ll n; scanf("%I64u",&n); printf("%I64u\n",(n+1)-solve(n+1)); } }
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