Question

我目前正在解决UVa Problem no. 10038。因此，为了调试我的代码，我去了uDebug。并发现我的代码几乎完整，但是有一些例外。

我希望查看以下代码。任何有关编辑的建议都可能会有所帮助。

我尝试使用while循环（在代码的近端）代替if语句，但这会冻结输入。

/*#include <iostream>, <vector>, <algorithm>, <functional>*/

int main(){
/*Declaring everything required*/
    int n, num, diff;
    bool is_jolly;
    vector<int> numbers, differences;

    while(cin >> n){
        /*Deleting all elements before working with them*/
        numbers.clear();
        differences.clear();

/*A sequence of single number is always JOLLY*/

            /*Taking in all the numbers*/
            for(int i = 0; i < n; i++){
                cin >> num;
                numbers.push_back(num);
            }

            /*Calculating the absolute value of those*/
            for(int i = 1; i < n; i++){
                diff = abs(numbers[i] - numbers[i - 1]);
                differences.push_back(diff);
            }

            /*Sorting the differences in descending order*/
            if(differences.size() > 1){
                sort(differences.begin(), differences.end(), greater<int>());
            }

            /*Creating a iterator for checking if the value reached one through n - 1*/
            int k = n - 1;
            for(int i = 0; i < differences.size(); i++){
                if(differences[i] == k){/*Tried the while loop here*/
                    is_jolly = true;/*<-----This is where the problem is. If the difference at first is equal to k then it will not check again. I want it to check as long as the loop runs*/
                }else{
                    is_jolly = false;
                }
            k--;
            }
/*if is_jolly is true will print "Jolly" else "Not jolly"*/

        }

    }
return 0;
}

输入为

1 2000
5 1 4 2 -1 6
2 1999 1998
4 1 4 2 3
4 1 3 2 -2 <---- Here I am getting Jolly instead of Not Jolly
4 1 4 3 5
4 1 2 5 7
3 4 1 3
4 1 4 2 3
4 1 2 3 6 <--- Here I am getting Jolly instead of Not Jolly
2 1 3
1 1

其他所有功能都按预期运行。

Answer 1

像这样更改此循环。

如果在任何时候，任何差异[i] 不等于 k ，那么它是不快活 >。否则，它是 jolly 。因此，我们将随时检查差异[i] 是否不等于k，然后 is_jolly 是 false 。

is_jolly = true; // suppose the answer is true
for(int i = 0; i < differences.size(); i++){
     if(differences[i] != k){    // if at any time this value is not k then answer is false.
          is_jolly = false;
          break;
     }
     k--;
}

任何人都可以查看以下UVa 10038解决方案代码吗？

1 个答案: