题意： 给你一个很大的数(1e5位），求不超过25个回文数和为这个数。
思路： 直接找尽量接近这个数的回文数，减去就好了。 收获是进一步熟悉了这个大整数板子（定义数组也是要时间的啊）
#pragma comment(linker, "/STACK:102400000,102400000") #include <iostream> #include <string> #include <cstring> #include <cstdio> #include <vector> using namespace std; const int maxn = 100004; struct bign{ int d[maxn], len; void clean() { while(len > 1 && !d[len-1]) len--; } bign() { memset(d, 0, sizeof(d)); len = 1; } bign(int num) { *this = num; } bign(char* num) { *this = num; } bign operator = (const char* num){ for(int i = 0;i < len;i++) d[i] = 0; // memset(d,0,sizeof(d)); len = strlen(num); for(int i = 0; i < len; i++) d[i] = num[len-1-i] - '0'; clean(); return *this; } bign operator = (int num){ char s[20]; sprintf(s, "%d", num); *this = s; return *this; } bign operator + (const bign& b){ bign c = *this; int i; for (i = 0; i < b.len; i++){ c.d[i] += b.d[i]; if (c.d[i] > 9) c.d[i]%=10, c.d[i+1]++; } while (c.d[i] > 9) c.d[i++]%=10, c.d[i]++; c.len = max(len, b.len); if (c.d[i] && c.len <= i) c.len = i+1; return c; } bign operator - (const bign& b){ bign c = *this; int i; for (i = 0; i < b.len; i++){ c.d[i] -= b.d[i]; if (c.d[i] < 0) c.d[i]+=10, c.d[i+1]--; } while (c.d[i] < 0) c.d[i++]+=10, c.d[i]--; c.clean(); return c; } bign operator * (const bign& b)const{ int i, j; bign c; c.len = len + b.len; for(j = 0; j < b.len; j++) for(i = 0; i < len; i++) c.d[i+j] += d[i] * b.d[j]; for(i = 0; i < c.len-1; i++) c.d[i+1] += c.d[i]/10, c.d[i] %= 10; c.clean(); return c; } bign operator / (const bign& b){ int i, j; bign c = *this, a = 0; for (i = len - 1; i >= 0; i--) { a = a*10 + d[i]; for (j = 0; j < 10; j++) if (a < b*(j+1)) break; c.d[i] = j; a = a - b*j; } c.clean(); return c; } bign operator % (const bign& b){ int i, j; bign a = 0; for (i = len - 1; i >= 0; i--) { a = a*10 + d[i]; for (j = 0; j < 10; j++) if (a < b*(j+1)) break; a = a - b*j; } return a; } bign operator += (const bign& b){ *this = *this + b; return *this; } bool operator <(const bign& b) const{ if(len != b.len) return len < b.len; for(int i = len-1; i >= 0; i--) if(d[i] != b.d[i]) return d[i] < b.d[i]; return false; } bool operator >(const bign& b) const{return b < *this;} bool operator<=(const bign& b) const{return !(b < *this);} bool operator>=(const bign& b) const{return !(*this < b);} bool operator!=(const bign& b) const{return b < *this || *this < b;} bool operator==(const bign& b) const{return !(b < *this) && !(b > *this);} string str() const{ string s; for(int i = len - 1; i >= 0; i--) s += d[i]+'0'; return s; } }; istream& operator >> (istream& in, bign& x) { string s; in >> s; x = s.c_str(); return in; } ostream& operator << (ostream& out, const bign& x) { out << x.str(); return out; } int main() { ios::sync_with_stdio(false); int T;cin >> T; bign num,now; while(T--) { cin >> num; int cnt = 30; vector<string>vec; while(cnt--) { if(num == 0) break; if(num.len == 1) { vec.push_back(num.str()); break; } now = num; int pos = 0; for(int i = now.len / 2;i < now.len;i++) { if(now.d[i] != 0) { now.d[i]--; pos = i; break; } } if(pos == now.len - 1) { if(num.d[num.len - 1] == 1) { now.len--; for(int i = 0;i < num.len - 1;i++) { now.d[i] = 9; } } else { now.d[num.len - 1] = num.d[num.len - 1] - 1; for(int i = 1;i < num.len - 1;i++) { now.d[i] = 9; } now.d[0] = num.d[num.len - 1] - 1; } } else { for(int i = 0;i < now.len / 2;i++) { now.d[i] = now.d[now.len - 1 - i]; } } vec.push_back(now.str()); num = num - now; } cout << vec.size() << endl; for(int i = 0;i < vec.size();i++) { cout << vec[i] << endl; } } return 0; }