TopCoder SRM606 DIV1 250

来源：互联网发布：pro怎么拆卸软件编辑：程序博客网时间：2024/05/17 22:51

题目大意

玩家2心里想一个数N(1≤N≤1,000,000,000),玩家1会进行Q(1≤Q≤50)次询问,每一次询问guess_i,玩家2会告诉玩家1:answers_i,其中answers_i=abs(guess_i-N).如果能通过这Q次询问猜出N,则输出N的大小.如果玩家2撒谎了,则输出-2,如果不能猜出N,则输出-1.

题解

涉及的数字为guess_i±answers_i且数目不会超过Q*2个我们可以用map来存下每一个数被数了几次,如果有有且只有一个数被数了Q次那么这个数就是答案,如果有多个数被数了Q次则输出-1,其他情况输出-2就好了.

代码

#include <vector>#include <map>#include <algorithm>#include <iostream>#include <cstdio>using namespace std;map<int,int> map1;class EllysNumberGuessing {public:    int getNumber(vector<int> guesses, vector<int> answers) {        int n = (int) guesses.size();        int flag = 0, mark = 0;        for (int i = 0; i < n; ++i) {            flag = 0;            if (guesses[i] - answers[i] >= 1) {                map1[guesses[i] - answers[i]]++;                if (map1[guesses[i] - answers[i]] == i + 1) {                    flag++;                    mark = guesses[i] - answers[i];                }            }            if (guesses[i] + answers[i] <= 1000000000LL) {                map1[guesses[i] + answers[i]]++;                if (map1[guesses[i] + answers[i]] == i + 1) {                    flag++;                    mark = guesses[i] + answers[i];                }            }        }        if (flag == 1) {            return mark;        }        if (!flag) {            return -2;        }        return -1;    }};

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