[编程题] 最大和
在一个N*N的数组中寻找所有横,竖,左上到右下,右上到左下,四种方向的直线连续D个数字的和里面最大的值
输入描述:
每个测试输入包含1个测试用例,第一行包括两个整数 N 和 D :
3 <= N <= 100
1 <= D <= N
接下来有N行,每行N个数字d:
0 <= d <= 100
输出描述:
输出一个整数,表示找到的和的最大值
输入例子:
4 2
87 98 79 61
10 27 95 70
20 64 73 29
71 65 15 0
输出例子:
193
代码
#include#include using namespace std; int main(void) { int row, num; cin >> row >> num; int input[100][100]; /* if (num > row) { return 0; } else { int sum = row*row; int **input = (int**)malloc(row*sizeof(int*)); for (int i = 0; i < row; i++) { input[i] = new int[row]; for (int j = 0; j < row; j++) { int cur; cin >> cur; input[i][j] = cur; } } */ for(int i = 0;i >d; input[i][j]=d; } } int maxsum = -1; int rownum, colnum, cnt, pos, cursum; //所有水平方向的和 for (rownum = 0; rownum < row; rownum++) { for (colnum = 0; colnum <= row - num; colnum++) { cnt = num; pos = colnum; cursum = 0; for (; cnt>0; cnt--) { cursum += input[rownum][pos]; pos++; } if (cursum > maxsum) maxsum = cursum; } } //所有垂直方向的和 for (colnum = 0; colnum < row; colnum++)//每一列 { for (rownum = 0; rownum <= row - num; rownum++) { cnt = num; pos = rownum; cursum = 0; for (; cnt>0; cnt--) { cursum += input[pos][colnum]; pos++; } if (cursum > maxsum) maxsum = cursum; } } //所有主对角线的和 for (rownum = 0; rownum <= row - num; rownum++) { for (colnum = 0; colnum < row - num; colnum++) { cnt = num; pos = 0; cursum = 0; for (; cnt>0; cnt--) { cursum += input[rownum + pos][colnum + pos]; pos++; } if (cursum > maxsum) maxsum = cursum; } } //所有次对角线的和 for (rownum = 0; rownum <= row - num; rownum++) { for (colnum = num - 1; colnum < row; colnum++) { cnt = num; pos = 0; cursum = 0; for (; cnt>0; cnt--) { cursum += input[rownum + pos][colnum - pos]; pos++; } if (cursum > maxsum) maxsum = cursum; } } cout << maxsum; return 0; //} //return 0; }
您的代码已保存
答案错误:您提交的程序没有通过所有的测试用例
case通过率为98.00%
测试用例:
23 14
对应输出应该为:
1017
你的输出为:
986