山东大学23强基班计算机导论第七次习题答案是什么？
摘要：T1.1 int day_of_year(int year,int month,int day) { static int sum[] = {0,31,28,31,30,31,30,31,31,30,31,30,31}; if(year %
T1.1
int day_of_year(int year,int month,int day)
{
static int sum[] = {0,31,28,31,30,31,30,31,31,30,31,30,31};
if(year % 4 == 0 && year % 100 != 0)
sum[2]++;
else if(year % 4 == 0 && year % 100 ==0 & year %400==0)
sum[2]++;
int offset = 0;
for(int i=1;i<month;i++)
offset+=sum[i];
offset+=day;
return offset;
}
T2.1
#include <stdio.h>
int main()
{
int sum = 0;
int n;
scanf("%d",&n);
for(int i = 0;i < n;i++)
for(int j = 0;j < n;j++)
{
int t;
scanf("%d",&t);
if(i+j==n-1 || i==n-1 || j==n-1)
continue;
sum+=t;
}
printf("%d",sum);
}
T2.2
#include <stdio.h>
int main()
{
int m,n;
scanf("%d%d",&m,&n);
for(int i=0;i<m;i++){
int sum=0;
for(int j=0;j<n;j++){
int t;
scanf("%d",&t);
sum+=t;
}
printf("%d\n",sum);
}
}
T2.3
#include <stdio.h>
#define N 128
int A[N][N];
int B[N][N];
int C[N][N];
int main()
{
int m1,n1,m2,n2;
scanf("%d%d",&m1,&n1);
for(int i=0;i<m1;i++)
for(int j=0;j<n1;j++)
scanf("%d",&A[i][j]);
scanf("%d%d",&m2,&n2);
for(int i=0;i<m2;i++)
for(int j=0;j<n2;j++)
scanf("%d",&B[i][j]);
if(n1!=m2)
{
printf("Error: %d != %d",n1,m2);
return 0;
}
// In this way,It`s will be faster than normal by about 4 times.
for(int k=0;k<n1;k++)
for(int i=0;i<m1;i++)
for(int j=0;j<n2;j++)
C[i][j]+=A[i][k]*B[k][j];
printf("%d %d\n",m1,n2);
for(int i=0;i<m1;i++){
for(int j=0;j<n2;j++){
printf("%d",C[i][j]);
if(j!=n2-1)
putchar(' ');
}
putchar('\n');
}
}
T2.4
#include <stdio.h>
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n;
scanf("%d",&n);
int flag = 1; // As a result of buffer.
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
{
int t;
scanf("%d",&t);
if(t!=0 && i>j){
// printf("%d %d %d\n",i,j,t);
flag=0;
}
}
if(flag)
puts("YES");
else
puts("NO");
}
}
T2.5
#include <stdio.h>
int main()
{
int m,n;
int maxn = -0x01145140;
int nx,ny;
scanf("%d%d",&m,&n);
for(int i=0;i<m;i++)
for(int j=0;j<n;j++)
{
int t;
scanf("%d",&t);
if(t>maxn)
{
maxn=t;
nx=i;ny=j;
}
}
printf("%d\n%d %d",maxn,nx,ny);
}
T2.6
#include <stdio.h>
int A[8][8];
int main()
{
int n;
scanf("%d",&n);
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
scanf("%d",&A[j][i]);
for(int i=0;i<n;i++,putchar('\n'))
for(int j=0;j<n;j++)
printf("%4d",A[i][j]);
// So easy.
}
T2.7
#include <stdio.h>
int A[12][12];
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
scanf("%d",&A[i][j]);
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
{
int t;scanf("%d",&t);
if(t!=A[i][j])
{
puts("No");
return 0;
}
}
puts("Yes");
}
T2.8
#include <stdio.h>
int A[12][12];
int main()
{
int n;
scanf("%d",&n);
int dx[]={1,-1,0,0};
int dy[]={0,0,1,-1};
// I think it have a math way,But A it will be faster to write.
int T[]={2,3,1,0};//trans matrix
int nx=0,ny=0,step=n*n,cur=1,way=0;
int a=0,b=n-1,c=1,d=n-1; // bound
while(step--)
{
A[ny][nx]=cur;
if(way==0 && nx==b){
way=T[way];
b--;
}
else if(way==1 && nx==a){
way=T[way];
a++;
}
else if(way==2 && ny==d){
way=T[way];
d--;
}
else if(way==3 && ny==c){
way=T[way];
c++;
}
nx+=dx[way];
ny+=dy[way];
cur++;
}
// Too ugly
for(int i=0;i<n;i++,putchar('\n'))
for(int j=0;j<n;j++)
printf("%3d",A[i][j]);
}
T2.9
#include <stdio.h>
int main()
{
int n;
scanf("%d",&n);
for(int i=0;i<n;i++,putchar('\n'))
{
for(int j=0;j<n;j++)
{
putchar('0' + (i==j));
putchar(' ');
}
}
}