topcoder SRM 625 DIV2 A

題目

Problem Statement
	You are given an int y. We are looking for any vector <int> x          that satisfies the following constraints:           x has exactly three elements ( x[0] * x[1] ) + x[2] = y Each x[i] must be between -1000 and 1000, inclusive. No x[i] can be equal to 0 or 1. Find and return one such x. If there are multiple          valid solutions, you may return any of them. You may assume that for          our constraints on y (specified below) at least one valid x          always exists.
Definition
	Class: AddMultiply Method: makeExpression Parameters: int Returns: vector <int> Method signature: vector <int> makeExpression(int y) (be sure your method is public)
Limits
	Time limit (s): 2.000 Memory limit (MB): 256
Constraints
-	y will be between 0 and 500, inclusive.
Examples
0)
	6                     Returns: {2, 2, 2 }               2*2 + 2 = 6 Note that this is one of many                      possible solutions. Another solution is: 3*3 +                      (-3) = 6
1)
	11                     Returns: {2, 3, 5 }
2)
	0                     Returns: {7, 10, -70 }               Note that 0 and 1 are not allowed, thus a result like 0                      * 0 + 0 would be incorrect.
3)
	500                     Returns: {-400, -3, -700 }               Some or all of the returned numbers may be negative.
4)
	2                     Returns: {2, 2, -2 }
5)
	5                     Returns: {5, 2, -5 }

This problem statement is the exclusive and proprietary property of      TopCoder, Inc. Any unauthorized use or reproduction of this information      without the prior written consent of TopCoder, Inc. is strictly      prohibited. (c)2003, TopCoder, Inc. All rights reserved.

分析

從反面，即何時不成立來思考

代碼

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;
int aPA[3] = { 2, 2, 0};
int aPB[3] = { 3, 3, 0};
vector<int> PA(aPA, aPA + 3);
vector<int> PB(aPB, aPB + 3);
#define PAPOS 0
#define NOPA1 4
#define NOPA2 5
#define PBPOS 1
#define NOPB1 9
#define NOPB2 10
#define POSMAX 2
#define INIT_ANS()    do{ ansTable[PAPOS] = PA; ansTable[PBPOS] = PB; }while(0)
#define MAXN 1000
#define MINN -1000
vector<int> ansTable[POSMAX];
class AddMultiply {
public:
 vector <int> makeExpression(int);
};
vector <int> AddMultiply::makeExpression(int y) {
    INIT_ANS();
    if(y != NOPA1 && y != NOPA2){
        ansTable[PAPOS][2] = y - 4;
        return ansTable[PAPOS];
    }else{
        ansTable[PBPOS][2] = y - 9;
        return ansTable[PBPOS];
    }
}