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

#include "bits/stdc++.h"
using namespace std;
const int MAX=1e5+5;
int t,ls1,ls2;
char s1[MAX],s2[MAX];
vector <string> ans;
int main(){
    freopen ("i.in","r",stdin);
    freopen ("i.out","w",stdout);
    int i,j,mid;
    scanf("%d",&t);
    while (t--){
        scanf("%s",s1+1);
        ls1=strlen(s1+1);
        reverse(s1+1,s1+ls1+1);
        ans.clear();
        while (ls1){
            if (ls1==2 && s1[2]=='1' && s1[1]=='0'){
                ans.push_back("1");
                ans.push_back("9");
                break;
            }
            if (ls1==1){
                ans.push_back(string(1,s1[1]));
                break;
            }
            mid=ls1/2;
            memset(s2,0,sizeof(s2));
            for (i=ls1;i>mid;i--) s2[i]=s1[i];
            s2[mid]--;
            i=mid;
            while (s2[i]<'0'){
                s2[i]+=10;
                s2[i+1]--;
                i++;
            }
            if (s2[ls1]=='0'){
                for (i=1;i<=mid;i++) s2[i]=s2[ls1-1-i+1];
                if (ls1%2==0) s2[mid]='9';
                ans.push_back(string(s2+1,s2+ls1));
            }
            else{
                for (i=1;i<=mid;i++) s2[i]=s2[ls1-i+1];
                ans.push_back(string(s2+1,s2+ls1+1));
            }
            for (i=1;i<=ls1;i++){
                s1[i]=s1[i]-(s2[i]-'0');
                if (s1[i]<'0'){
                    s1[i]+=10;
                    s1[i+1]--;
                }
            }
            while (s1[ls1]=='0' && ls1) ls1--;
        }
        printf("%d
",ans.size());
        for (i=0;i<ans.size();i++)
            cout<<ans[i]<<endl;
    }
    return 0;
}