HUST 1599 Multiple

Multiple

Description

Rocket323 loves math very much. One day, Rocket323 got a number
string. He could choose some consecutive digits from the string to form a
number. Rocket323loves 64 very much, so he wanted to know how
many ways can he choose from the string so the number he got multiples of 64
?
 
A number cannot have leading zeros. For example, 0, 64, 6464, 128
are numbers which multiple of 64 , but 12, 064, 00, 1234 are not.

Input

Multiple cases, in each test cases there is only one line consist a number
string.
Length of the string is less than 3 * 10^5 .
 
Huge Input
, scanf is
recommended.

Output

Print the number of ways Rocket323 can
choose some consecutive digits to form a number which multiples of 64.

Sample Input

64064

Sample Output

5

HINT

There are five substrings which multiples of
64.
[64]064
640[64]
64[0]64
[64064]
[640]64

题目大意是：给出一个串，求有多少个子串（无前导0）能被64整除。

分析：单独的一个子串“0”肯定是的，再暴搜子串字符个数为2、3、4、5、6的能被64整除的，1000000%64==0。

所以字符个数为6的可以整除64的可以特殊判断一下：

高位为0，那么他本身不符合（有前导0），但在它前面加任意一个不为0的数都能整除64。所以加上他前面不为0的数字个数。

高位不为0，那么再加上它本身这一种。

 1 #include<iostream>

 2 #include<cstring>

 3 #include<cstdio>

 4 using namespace std;

 5

 6 typedef __int64 LL;

 7 const int maxn=300005;

 8 char text[maxn];

 9 int d[maxn];

10

11 void solve()

12 {

13     int i,len=strlen(text);

14     if(text[0]==‘0‘) d[0]=1;

15     else d[0]=0;

16     for(i=1;i<len;i++)

17     {

18         if(text[i]==‘0‘)  d[i]=d[i-1]+1;

19         else d[i]=d[i-1];

20     }

21     LL ans=d[len-1];

22     for(i=len-1;i>=1;i--)

23         if(text[i-1]!=‘0‘ && ((text[i-1]-‘0‘)*10+text[i]-‘0‘)%64==0) ans++;

24     for(i=len-1;i>=2;i--)

25         if(text[i-2]!=‘0‘ && ((text[i-2]-‘0‘)*100+(text[i-1]-‘0‘)*10+text[i]-‘0‘)%64==0) ans++;

26     for(i=len-1;i>=3;i--)

27         if(text[i-3]!=‘0‘ && ((text[i-3]-‘0‘)*1000+(text[i-2]-‘0‘)*100+(text[i-1]-‘0‘)*10+text[i]-‘0‘)%64==0) ans++;

28     for(i=len-1;i>=4;i--)

29         if(text[i-4]!=‘0‘ && ((text[i-4]-‘0‘)*10000+(text[i-3]-‘0‘)*1000+(text[i-2]-‘0‘)*100+(text[i-1]-‘0‘)*10+text[i]-‘0‘)%64==0) ans++;

30     for(i=len-1;i>=5;i--)

31     {

32         if(((text[i-5]-‘0‘)*100000+(text[i-4]-‘0‘)*10000+(text[i-3]-‘0‘)*1000+(text[i-2]-‘0‘)*100+(text[i-1]-‘0‘)*10+text[i]-‘0‘)%64==0)

33         {

34             if(text[i-5]!=‘0‘) ans++;

35             if(i-5>0) ans+=i-5-d[i-5-1];

36         }

37     }

38     printf("%I64d\n",ans);

39 }

40 int main()

41 {

42     while(gets(text))

43     {

44         solve();

45     }

46     return 0;

47 }

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