暑假集训Day22 I (模拟+高精度)

题目链接在这里：200202.pdf (codeforces.com)

其实思路非常好想，就是每次折半，把前一半回文到后面，这样的复杂度是log肯定可以行得通。然后如果最后剩的是10的话要特判一下

需要注意的是如果在结构体里面开很大的数组比如1e5以上的话会很有可能RE

 1 #include "bits/stdc++.h"
 2 using namespace std;
 3 const int MAX=1e5+5;
 4 int t,ls1,ls2;
 5 char s1[MAX],s2[MAX];
 6 vector <string> ans;
 7 int main(){
 8     freopen ("i.in","r",stdin);
 9     freopen ("i.out","w",stdout);
10     int i,j,mid;
11     scanf("%d",&t);
12     while (t--){
13         scanf("%s",s1+1);
14         ls1=strlen(s1+1);
15         reverse(s1+1,s1+ls1+1);
16         ans.clear();
17         while (ls1){
18             if (ls1==2 && s1[2]=='1' && s1[1]=='0'){
19                 ans.push_back("1");
20                 ans.push_back("9");
21                 break;
22             }
23             if (ls1==1){
24                 ans.push_back(string(1,s1[1]));
25                 break;
26             }
27             mid=ls1/2;
28             memset(s2,0,sizeof(s2));
29             for (i=ls1;i>mid;i--) s2[i]=s1[i];
30             s2[mid]--;
31             i=mid;
32             while (s2[i]<'0'){
33                 s2[i]+=10;
34                 s2[i+1]--;
35                 i++;
36             }
37             if (s2[ls1]=='0'){
38                 for (i=1;i<=mid;i++) s2[i]=s2[ls1-1-i+1];
39                 if (ls1%2==0) s2[mid]='9';
40                 ans.push_back(string(s2+1,s2+ls1));
41             }
42             else{
43                 for (i=1;i<=mid;i++) s2[i]=s2[ls1-i+1];
44                 ans.push_back(string(s2+1,s2+ls1+1));
45             }
46             for (i=1;i<=ls1;i++){
47                 s1[i]=s1[i]-(s2[i]-'0');
48                 if (s1[i]<'0'){
49                     s1[i]+=10;
50                     s1[i+1]--;
51                 }
52             }
53             while (s1[ls1]=='0' && ls1) ls1--;
54         }
55         printf("%d\n",ans.size());
56         for (i=0;i<ans.size();i++)
57             cout<<ans[i]<<endl;
58     }
59     return 0;
60 }

posted on 2021-08-25 20:42 珍珠鸟阅读(33) 评论(0) 编辑收藏举报