CSAcademy Prefix Suffix Counting 题解
table of Contents
The meaning of problems
Give you two numbers \ (N \) and \ (M \) , now \ (K \) represent (M \) \ digits. Ask you from \ (1 \) to \ (N \) , the number of digits meet \ (M \) at the same time its former \ (K \) numbers and the \ (K \) digits.
Thinking
We now assume \ (N \) and \ (M \) are strings, if not specifically mentioned, are treated as strings.
Suppose a digital \ (S \) greater than or equal \ (1 \) or less \ (N \) , we discuss its condition can be classified :( same \ (S \) as strings)
- \ (S \) greater than the length equal to \ (2K \) , then the prefix and suffix not affect each other, Motian part of the intermediate space in the figure is a value that satisfies less \ (N \) in the case of just fill .
- \ (S \) length is less than \ (2K \) , then the prefix and suffix overlap each other, you need to ensure that they do not conflict with each other, which is still to meet before the suffix is \ (M \) . And, you do not have digital free to go fill up.
practice
In the case mentioned above, may be implemented separately:
- We enumerate the string length of the current / number, if its total length is less than \ (N \) overall length, it must be smaller than \ (N \) , when the intermediate space can easily fill. If its total length is equal to (N \) \ lengths, then we enumerate a bit in this one before, all the numbers are and \ (N \) is the same, which is a digital than \ ( N \) of this one small, then just fill the back. In particular, it is possible that a previous \ (K \) appears numbers in the past, so in the middle of all the digits are easily filled, or the middle of all the digits are filled \ (0 \) , the current figure is still more than \ (N \) big, then there is no legitimate program.
- Similarly, the enumeration current string length / number of overlapping portions satisfy the prefix and suffix of the same conditions require rapid determination, may be used
ZorKMPother fast method for string matching.
program
With practice , before using Z matching suffix.
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=1000000007;
char s[1000005];//s表示题目中的数字N
char t[1000005];//t表示题目中的数字M
int n,m;//n, m分别表示题目中的N, M的长度
ll pw[1000005],ans;//pw[i]表示10的i次方对mod取模后的大小，ans表示最终答案
bool check(int emp){//check表示中间空出了emp个填0的数位的时候，前后缀是M的数字是否小于等于N，emp<0时，表示前后缀的M重叠了几个位置
char *r=new char[n+1];
memset(r,'0',n+1);
memcpy(r+1+n-(m+emp+m),t+1,m);
memcpy(r+1+n-m,t+1,m);//构造一个前后缀为M的数字，长度小于N时高位补零
bool fl=false,va=true;
for(int i=1;i<=n;i++){
if(r[i]<s[i])fl=true;
if(r[i]>s[i]&&!fl)va=false;
}//判断大小关系（数学意义）
delete[] r;
return va;
}
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
pw[0]=1;
for(int i=1;i<=1000000;i++)pw[i]=pw[i-1]*10%mod;//预处理
cin>>s+1;
n=strlen(s+1);
cin>>t+1;
m=strlen(t+1);
if(check(-m))ans++;//单个M作为数字，这里其实可能会出锅，但数据中没有
int *z=new int[m+1],l=1,r=1;
z[1]=m;
for(int i=2;i<=m;i++){
z[i]=0;
if(i>r){
l=i;r=i-1;
while(r<m&&t[r+1]==t[z[i]+1])r++,z[i]++;
}else{
if(i+z[i-l+1]<=r){
z[i]=z[i-l+1];
}else{
l=i;
z[i]=r-i+1;
while(r<m&&t[r+1]==t[z[i]+1])r++,z[i]++;
}
}
if(i+z[i]-1==m){
if(m+m-z[i]<n||(m+m-z[i]==n&&check(-z[i])))ans++;
}
}
delete[] z;//Z算法
if(m+m<n||(m+m==n&&check(0)))ans++;//计算有数字形如M+M时是否合法
for(int i=1;m+i+m<n;i++){//枚举空出了i个数字时
ans=(ans+pw[i])%mod;
}
if(m+m<n&&check(n-m-m)){
int emp=n-m-m;
bool pref=false,suff=true,suf=false;//pref表示前K个数字是否已经比N小了，这样可以随便填，suff表示M是否小于等于N的K个数字，是就可以让中间的数字全部等于N对应数位
for(int i=1;i<=m;i++){
if(s[i]>t[i])pref=true;
if(s[i+n-m]>t[i])suf=true;
if(s[i+n-m]<t[i]&&!suf)suff=false;
}
if(pref)ans=(ans+pw[emp])%mod;//加上中间任意填的方案
else{
ans=(ans+suff)%mod;//加上suff的方案
for(int i=m+1;i<=n-m;i++){
ans=(ans+pw[n-m-i]*(s[i]-'0'))%mod;//加上当前位前面的数字和N相同，当前位小于N对应数位，后面随便填的方案
}
}
}
cout<<ans%mod<<endl;
return 0;
}