I.[音频]Listen to the video clip and tell that the phrase“my head is pounding ”mean.
I.[音频]Listen to the video clip and tell that the phrase“my head is pounding ”mean.
中国大学MOOC: OPT[i][w]=max{OPT[i-1][w],OPT[i][w-w[i]]+v[i]},这是()问题的递推关系。[/i][/i][/i][/i]
中国大学MOOC: OPT[i][w]=max{OPT[i-1][w],OPT[i][w-w[i]]+v[i]},这是()问题的递推关系。[/i][/i][/i][/i]
#include [stdio.h]int main() { int a[3][3],i,j; for(i=0;i<3;i++) for(j=0;j<3;j++) a[i][j]=3*i+j; for(i=0;i<3;i++) printf("%d ",a[i][1]); return 0;}[/i][/i]
#include [stdio.h]int main() { int a[3][3],i,j; for(i=0;i<3;i++) for(j=0;j<3;j++) a[i][j]=3*i+j; for(i=0;i<3;i++) printf("%d ",a[i][1]); return 0;}[/i][/i]
下面代码的输出结果是: 。ls = [[1,3,6],[4,6,3],[2,8,5]]sum1,sum2 = 0,0for i in range(3):sum1 = sum1 + ls[i][i]sum2 = sum2 + ls[i][2-i]print("sum1={},sum2={}".format(sum1,sum2))[/i][/i][/i]
下面代码的输出结果是: 。ls = [[1,3,6],[4,6,3],[2,8,5]]sum1,sum2 = 0,0for i in range(3):sum1 = sum1 + ls[i][i]sum2 = sum2 + ls[i][2-i]print("sum1={},sum2={}".format(sum1,sum2))[/i][/i][/i]
若有int a[5][5];则*(a+i)+j是a[i][j]的地址,*(a+i)等价于a[i]等价于&a[i][0][/i][/i][/i]
若有int a[5][5];则*(a+i)+j是a[i][j]的地址,*(a+i)等价于a[i]等价于&a[i][0][/i][/i][/i]
已知列表m=[[1,2],[3,4]],有列表a=[[row[i] for row in m] for i in range(2)],则a[0][1]是[/i]
已知列表m=[[1,2],[3,4]],有列表a=[[row[i] for row in m] for i in range(2)],则a[0][1]是[/i]
已知x=[[1,2,3],[4,5,6]],表达式[[row[i] for row in x] for i in range(len(x[0]))]的值为(______ )。[/i]
已知x=[[1,2,3],[4,5,6]],表达式[[row[i] for row in x] for i in range(len(x[0]))]的值为(______ )。[/i]
n阶对称矩阵a满足a[i][j]=a[j][i],i,j=1…n,用一维数组t存储时,t的长度为____,当i=j,a[i][j]=t[2],i>;j,a[i][j]=t[3],i<;j,a[i][j]=t[4]。[/i][/i][/i][/i][/i]
n阶对称矩阵a满足a[i][j]=a[j][i],i,j=1…n,用一维数组t存储时,t的长度为____,当i=j,a[i][j]=t[2],i>;j,a[i][j]=t[3],i<;j,a[i][j]=t[4]。[/i][/i][/i][/i][/i]
print([i for i in range(11) if i >; 4])
print([i for i in range(11) if i >; 4])
以下程序的运行结果是#include <;stdio.h>;main(){ int i,j,x[3][3]={0}; for(i=0;i<;3;i++)for(j=0;j<;=i;j++)x[i][j]=i*j; printf("%d,%d",x[1][2],x[2][1]);}[/i]
以下程序的运行结果是#include <;stdio.h>;main(){ int i,j,x[3][3]={0}; for(i=0;i<;3;i++)for(j=0;j<;=i;j++)x[i][j]=i*j; printf("%d,%d",x[1][2],x[2][1]);}[/i]