2019 cattle off more training school fourth K.number (thinking)

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Meaning of the questions:

Enter a numeric string containing only obtained is a multiple of the number of substrings 300 (0,00,000 different positions so calculated, and consider the case of leading zeros).

sample input：

600

123000321013200987000789

sample output：

4

55

answer:

O (n) approach: iterate again, seeking to take the prefix and the sum of more than 3, the number of statistical sum of the num [sum], experience-based and next are 0, the statistics put the number before adding, finally add the number 0 alone.

O (300n) DP practices: the following

Official Solution:

Code：

O (n), the following:

 /*6ms*/
 1 #include<bits/stdc++.h>
 2 typedef long long ll;
 3 using namespace std;
 4 const int M=1e5+5;
 5 char s[M];
 6 int main()
 7 {
 8     while(~scanf("%s",s+1))
 9     {
10         int len=strlen(s+1),sum=0,num[3]={};
11         ll cnt=0;
12         num[0]=1;
13         for(int i=1;i<=len;i++){
14             sum=(sum+(s[i]-'0'))%3;
15             if(s[i]=='0'&&s[i+1]=='0'){
16                 cnt+=num[sum];
17             }
18             if(s[i]=='0')cnt++;
19             num[sum]++;
20         }
21         printf("%lld\n",cnt);
22     }
23     return 0;
24 }

 DP approach is as follows [O (300n)]:

 /*224ms*/
 1 #include<bits/stdc++.h>
 2 typedef long long ll;
 3 using namespace std;
 4 const int M=1e5+5;
 5 char s[M];
 6 int dp[M][302];
 7 int main()
 8 {
 9     while(~scanf("%s",s+1))
10     {
11         memset(dp,0,sizeof(dp));
12         dp[1][s[1]-'0']=1;
13         int len=strlen(s+1);
14         for(int i=2;i<=len;i++){
15             dp[i][s[i]-'0']++;
16             for(int j=0;j<300;j++){
17                 int flag=(j*10+s[i]-'0')%300;
18                 dp[i][flag]+=dp[i-1][j];
19             }
20         }
21         ll ans=0;
22         for(int i=1;i<=len;i++)
23             ans+=dp[i][0];
24         printf("%lld\n",ans);
25     }
26     return 0;
27 }