HDOJ 1554

You are to find all pairs of integers such that their sum is equal to the given integer number N and the second number results from the first one by striking out one of its digits. The first integer always has at least two digits and starts with a non-zero digit. The second integer always has one digit less than the first integer and may start with a zero digit.
Input
The input file consists of several test cases.
Each test case contains single integer N (10 ≤ N ≤ 10^9), as decribed above
Output
The output consists of several blocks, one for each test case.
On the first line of a block write the total number of different pairs of integers that satisfy the problem statement. On the following lines write all those pairs. Write one pair on a line in ascending order of the first integer in the pair. Each pair must be written in the following format: X + Y = N Here X, Y, and N, must be replaced with the corresponding integer numbers. There should be exactly one space on both sides of ‘+’ and ‘=’ characters.
Sample Input
302
Sample Output
5
251 + 51 = 302
275 + 27 = 302
276 + 26 = 302
281 + 21 = 302
301 + 01 = 302

hdu题解

试了一下，用哈希表去重在排序比set有序插入要快，插入查找哈希表得期望还是要比红黑树好一些。

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#include <iostream>
#include <algorithm>
#include <string>
#include <cmath>
#include<vector>
#include <unordered_set>
using namespace std;
int N;

unordered_set<int>Ans;

#define Verify verify(a,b,c,i)

inline bool verify(const int& a, const int& b, const int& c,const int&i) {
	return ((11 * a + b) * i + 2 * c) == N;
}

inline void UpdateAns(const int&X) {
	if (Ans.find(N - X) != Ans.end()) {
		return;
	}
	Ans.insert(X);
}

void Compute() {
	for (long long i = 1; i <= N; i *= 10) {
		int a = (N / i) / 11;
		int b = (N / i) % 11;
		int c;
		if (b < 10) {
			c = (N - N / i * i) / 2;
			if (Verify) {
				UpdateAns(10 * a * i + b * i + c);
			}
		}
		--b;
		if (a + b && b >= 0) {
			c = (N - N / i * i + i) / 2;
			if (Verify) {
				UpdateAns(10 * a * i + b * i + c);
			}
		}
	}
}

inline void OutputZero(int X) {

	int Tail = 0;
	int Y = N - X;
	while (X) {
		X /= 10;
		++Tail;
	}
	if (!Y) {
		--Tail;
	}
	while (Y) {
		Y /= 10;
		--Tail;
	}
	while (--Tail) {
		putchar('0');
	}
}

void Output(const vector<int>&Ans) {
	printf("%d\n", Ans.size());
	for (auto it = Ans.cbegin(); it != Ans.cend(); ++it) {
		printf("%d + ", *it);
		OutputZero(*it);
		printf("%d = %d\n", N - *it, N);
	}
}

int main(){

	while (cin >> N) {
		Compute();
		vector<int> ve;
		for (auto it = Ans.begin(); it != Ans.end(); ++it) {
			ve.push_back(*it);
		}
		sort(ve.begin(), ve.end(), [](int& x,int& y) {return x < y; });
		Output(ve);
		Ans.clear();
	}

	return 0;

}