说起这道题, 真是一把辛酸泪.
题意
将一个正整数 $n(le 10^{1000})$ 分解成不超过50个回文数的和.
做法
构造.
队友UHC提出的一种构造方法, 写起来比较方便一些, 而且比较巧妙. 可惜我码力太弱, 现场没调出来.
大体的想法是:
将一个数"等"分成两半. 如果长度为奇数可以采取下面两种处理方法之一.
- 在开头补0.
- 后一半 (低位) 的长度向下取整.
我采取第2种处理方式.
设前一半的长度为$h_1$, 后一半的长度为 $h_2$.
然后将前一半长
(写不下去了..., 贴代码...)
Implementation
#include <bits/stdc++.h>
using namespace std;
using num=vector<int>;
using iter=num::iterator;
using riter=num::reverse_iterator;
bool Less(const num &a, const num &b){ // a and b are of same length
for(auto it1=a.rbegin(), it2=b.rbegin(); it1!=a.rend(); ++it1, ++it2)
if(*it1 != *it2){
return *it1 < *it2;
}
return false;
}
void borrow(num &a){
for(int i=a.size()/2; i<a.size(); i++)
if(a[i]){
for(--a[i]; i; a[--i]+=9); //?
break;
}
while(a.size() && *a.rbegin()==0)
a.pop_back();
}
void split(num &lo, num &hi, const num &a){
auto mid=a.begin()+a.size()/2;
lo={a.begin(), mid};
hi={mid, a.end()};
}
num operator-(num a, num b){ // a and b are of same length && a >= b
num res;
for(int i=0; i<a.size(); i++){
if(a[i]<b[i]){
a[i]+=10;
a[i+1]-=1;
}
res.push_back(a[i]-b[i]);
}
for(; res.size() && *res.rbegin()==0; )
res.pop_back();
if(res.size()){ //error-prone
for(int i=0; i<res.size()-1; i++){
res[i+1]+=res[i]/10;
res[i]%=10;
}
for(; *res.rbegin()>9; ){
int t=*res.rbegin();
*res.rbegin()%=10;
res.push_back(t/10);
}
}
for(auto x:res){
assert(x>=0 && x<10);
}
return res;
}
vector<num> res;
void out(const num &a){
for(auto it=a.rbegin(); it!=a.rend(); ++it)
cout<<*it;
cout<<endl;
}
void solve(num &a){
res.clear();
num lo, hi;
int ones=0;
while(a.size()>1){
// out(a);
split(lo, hi, a);
num rhi(hi.rbegin(), hi.rbegin()+lo.size());
if(Less(lo, rhi)){
borrow(a);
++ones;
continue;
}
a=lo-rhi;
// There is no advantage to using {} initialization, and one trap,
// when using auto to get the type determined by the initializer.
// The trap is that if the initializer is a {}-list, we may not want its
// type deduced.
//initializer_list<num> xx={rhi};
//cout<<typeid(xx).name()<<endl;
auto tmp=rhi;
//cout<<typeid(tmp).name()<<endl;
for(auto &x: hi)
tmp.push_back(x);
res.push_back(tmp);
}
if(a.size()) res.push_back(a);
assert(res.size() + ones <= 50);
printf("%d
", res.size()+ones);
for(auto &x: res)
out(x);
while(ones--)
puts("1");
}
char s[1<<10];
int main(){
// num x, y;
// in g++:
// direct-list-intialization of 'auto' requires exactly one element
// for deduction to 'std::initializer_list', use copy-list-intialization (i.e. add '=' before the '{')
// auto xx{x, y};
int T, cas=0;
num a;
for(cin>>T; T--; ){
scanf("%s", s);
a.clear();
int n=strlen(s);
for(int i=n-1; i>=0; --i)
a.push_back(s[i]-'0');
printf("Case #%d:
", ++cas);
solve(a);
}
return 0;
}