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...main() {int i,j,sum; sum=0; for(i=1;i<10:i++) {for(j=1;j<10...
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Answer to main() { int array[10][5] = {0}; int i, j, sum = 0;
Yes. In the first example, the i loop runs N times, and the inner j loop tuns i*i times, which is O(N^2).So the whole thing is O(N^3).. In the second example there is an additional O(N^4) loop (loop to j*j), so it is O(N^5) overall.. For a more formal proof, work out how many times sum++ is executed in terms of N, and look at the highest polynomial order of N.
As a result, the line sum += 1; is executed only once outside of the loop, resulting in sum being incremented by 1 only a single time. Hence, after the execution of the code, the value of sum will be 1, which matches option C.
这段 C 语言代码是一个简单的程序，其目的是计算从 1 到 100 的整数之和，并将结果打印出来。让我们逐行分析： ```c #include // 引入标准输入输出库，通常用于处理文件和数据输出 int main() // 主函数，程序的入口点 { int i, sum = 0; // 定义两个变量，i 用于循环计数，sum 存储累加和初始值为 0 i = 1 ...
void main() {int i,j,sum; sum=0; for(i=1;i<10:i++) {for(j=1;j<10:j++) {sum+=i*j; } } printf过程都打印出来了#includeint main(){ int i,j,sum; sum=0 ...
int i,j,sum=0首先是声明和初始化 然后进入for语句，首先赋初值，i=1，j=10 然后判断i<10是否成立 因为1<10 所以执行循环语句sum=sum+i+j
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Question Time complexity of the following code snippet -: sum = 0; for (int i = 0; i < n; i++) for (int j = 0; j < i*i; j++) for (int k = 0; k < j; k++) sum++; My ...
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What is the output of the code fragment given below? int i = 0; int j = 0; while (i < 125) {i = i + 2; j++;} System.out.println(j); 63 What is the output of the following code fragment? int i = 1; int sum = 0; while (i <= 15) { sum = sum + i; i++; } System.out.println("The value of sum is " + sum);