大体题意：
给你一个长度不超过1000的大数，划分成n 个回文串，使得回文串的个数不超过50，并且回文串的和是这个大数！
思路：
大数模拟！
最长长度为1000 个数不超过50，显然必须是log级别的，也就是接近折半的运算！
所以我们先根据这个大数的前一半算出这个回文串是多少来，如果这个回文串小于当前的大数，直接减，在继续递归寻找！
如果大于当前大数， 你就必须找到一个非常接近它的回文串，其实也很好想，找到前一半，让前一半减1，这样在构造出的回文串一定小于当前大数，在继续递归即可！
比如说34567 直接构造34543 递归即可！
如果是 34511 构造出来34543 不能继续减，就让345减1，得344，构造出来的是34443 可以继续减，并且接近折半运算！
教训：
这种恶心的题目应对起来 可以写几个debug函数 看看哪里错了，这样更容易查错！
详细见代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <vector>
#include <assert.h>
using namespace std;
const int maxn = 2000 + 1;
vector<string>sans;
int cmpe(string s1,string s2){ // s1 > s2 = 1 s1 = s2 = 0 s1 < s2 = -1
int len1= s1.length();
int len2= s2.length();
if (len1 > len2) return 1;
else if (len1 < len2) return -1;
else {
if (s1 > s2) return 1;
else if (s1 < s2) return -1;
return 0;
}
}
bool judge(string s){
int len = s.length();
for (int i = 0; i < (len+1)/2; ++i){
if (s[i] != s[len-i-1])return false;
}
return true;
}
/**
1111111111111111110
998877665544332211
01122334455667789/9
**/
string Mul(string s1,string s2){
int len1 = s1.length();
int len2 = s2.length();
string ans = "";
int tmp = len1-len2;
int flag = 0;
for (int i = len2-1; ~i; --i){
if (s1[i+tmp]-flag >= s2[i]){
ans = char(s1[i+tmp] - s2[i] - flag + 48) + ans;
flag = 0;
}
else{
ans = char(s1[i+tmp] + 10 - s2[i] - flag + 48) + ans;
flag = 1;
}
}
for (int i = tmp-1; ~i; --i){
if (s1[i] - 48 - flag >= 0) ans = char(s1[i] - flag) + ans,flag = 0;
else ans = char(s1[i] - flag + 10) + ans,flag = 1;
}
int p = 0;
int anslen = ans.length();
while(p < anslen && ans[p] == 48)++p;
if (p != 0){
if (p == anslen){
ans[0] = 48;
ans[1] = 0;
ans.resize(1);
return ans;
}
for (int i = 0; i < anslen - p; ++i){
ans[i] = ans[i+p];
}
ans.resize(anslen-p);
}
return ans;
}
string unit = "1";
void deal(string s){
if (judge(s)){
sans.push_back(s);
return;
}
int len = s.length();
string t1;
t1.resize(len);
for(int i = 0; i < ((len&1)?len/2+1:len/2); ++i){
t1[i] = t1[len-i-1] = s[i];
}
string t2;
if (cmpe(s,t1) != -1){
sans.push_back(t1);
t2 = Mul(s,t1);
deal(t2);
}
else {
bool ok = 0;
for (int i = 1; i < len-1; ++i) if (t1[i] != '0'){
ok = 1;
break;
}
if (!ok && t1[0] == t1[len-1] && t1[0] == 49){
string zha = "1";
sans.push_back(zha);
zha = "";
// puts("1");
for (int i = 1; i < len; ++i) zha += "9";
sans.push_back(zha);
return;
}
t1.resize(len+1>>1);
t2 = Mul(t1,unit);
t2.resize(len);
for (int i = 0; i < len/2; ++i){
t2[len-i-1] = t2[i];
}
sans.push_back(t2);
t1 = Mul(s,t2);
deal(t1);
}
}
char s[maxn];
void print(vector<string>& aa){
printf("%d\n",aa.size());
for (int i = 0; i < aa.size(); ++i)printf("%s\n",aa[i].c_str());
}
int main(){
// debug();
int T,ks = 0;
scanf("%d",&T);
while(T--){
sans.clear();
scanf("%s",s);
printf("Case #%d:\n",++ks);
deal(s);
print(sans);
}
return 0;
}
Ugly Problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 898 Accepted Submission(s): 311
Special Judge
Problem Description
Everyone hates ugly problems.
You are given a positive integer. You must represent that number by sum of palindromic numbers.
A palindromic number is a positive integer such that if you write out that integer as a string in decimal without leading zeros, the string is an palindrome. For example, 1 is a palindromic number and 10 is not.
Input
In the first line of input, there is an integer T denoting the number of test cases.
For each test case, there is only one line describing the given integer s (
1≤s≤101000).
Output
For each test case, output “Case #x:” on the first line where x is the number of that test case starting from 1. Then output the number of palindromic numbers you used, n, on one line. n must be no more than 50. en output n lines, each containing one of your palindromic numbers. Their sum must be exactly s.
Sample Input
2
18
1000000000000
Sample Output
Hint
Source
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