1. Enter two sets of time (hours h, minutes m, seconds s) and calculate the average time. The difference between the two sets of times shall not exceed 1 hour, and h is between 0-11.
Input sample
1 20 30 1 30 30
0 20 30 11 30 30
Output
1 25 30
11 55 30
Reference code:
Method 1:
#include<stdio.h>
int main()
{
int a[2][2];
int i,j;
int h,m,s;
for(i=0;i<2;i++)
{
for(j=0;j<2;j++)
{
scanf("%d%d%d",&h,&m,&s);
if(h==0)
h=12;
a[i][j]=h*3600+m*60+s;
}
}
int sum[2];
sum[0]=(a[0][0]+a[0][1])/2;
sum[1]=(a[1][0]+a[1][1])/2;
for(i=0;i<2;i++)
{
printf("%d %d %d",sum[i]/3600,sum[i]%3600/60,sum[i]%60);
printf("\n");
}
}
Method Two:
#include <stdio.h>
int main()
{
int h1,h2,m1,m2,s1,s2;
int h,m,s,sum1,sum2,sum;
while(~scanf("%d%d%d%d%d%d",&h1,&m1,&s1,&h2,&m2,&s2))
{
if(h1==0) h1=12;
if(h2==0) h2=12;
sum1=h1*60*60+m1*60+s1;
sum2=h2*60*60+m2*60+s2;
sum=sum1+sum2;
sum=sum/2;
h=sum/3600%12;
m=(sum-h*3600)/60;
s=sum-h*3600-m*60;
printf("%d %d %d",h,m,s);
printf("\n");
}
}
2. Sort: input n, there are n groups of test numbers, each group has a variable length, and the output is sorted from small to large.
Input example:
2
1 4 7 2 5
8 9 4 3
Output:
1 2 4 5 7
3 4 8 9
Reference Code:
#include<stdio.h>
#include<string.h>
sort(int a[],int n)
{
int i,j,k;
for(i=0;i<n-1;i++)
{
for(j=0;j<n-i-1;j++)
{
if(a[j]>a[j+1])
{
k=a[j];
a[j]=a[j+1];
a[j+1]=k;
}
}
}
}
int main()
{
int i,n,len,a[100];
char ch;
scanf("%d%*c",&n);
while(n--)
{
len=0;
do
{
scanf("%d",&a[len]);
len++;
}while((ch=getchar())!='\n');
sort(a,len);
for(i=0;i<len;i++)
printf("%d ",a[i]);
printf("\n");
}
}
3. Enter the number of rows, and then enter an expression in each row to get the result.
Input example:
3
1+1
2.2/3
1+2*3
Output:
2
0.7
7
Reference Code:
#include<stdio.h>
#define StackSize 100
#define QueueSize 100
//队的DS定义
typedef struct
{
char data[100];
int front,rear; //front -- 头指针 rear -- 尾指针
}SeqQueue;
//队的初始化
void InitQueue(SeqQueue *Q)
{
Q->front = 0;
Q->rear = 0;
}
//队的判空 1-空 0-没空
int QueueEmpty(SeqQueue Q)
{
if(Q.front==Q.rear)
return 1;
else
return 0;
}
//入队
int EnQueue(SeqQueue *Q,char x)
{
if((Q->rear+1) % QueueSize == Q->front)
return 0;
else
{
Q->data[Q->rear] = x;
Q->rear = (Q->rear+1) % QueueSize;
}
}
//出队
char DeQueue(SeqQueue *Q)
{
char x;
if(Q->rear == Q->front)
return 0;
else
{
x = Q->data[Q->front];
Q->front = (Q->front + 1) % QueueSize;
return x;
}
}
//栈的DS定义
typedef struct
{
char data[100];
int top;
}SeqStack;
//栈的初始化
void InitStack(SeqStack *S)
{
S->top = -1;
}
//入栈
int Push(SeqStack *S,char x)
{
if(S->top == StackSize - 1)
return 0;
else
{
S->top++;
S->data[S->top] = x;
}
}
//出栈
char Pop(SeqStack *S)
{
char x;
if(S->top == -1)
return 0;
else
{
x = S->data[S->top];
S->top--;
return x;
}
}
//取栈顶元素
char GetTop(SeqStack S)
{
char x;
x = S.data[S.top];
return x;
}
//判断栈空 1-空 0-没空
int EmptyStack(SeqStack S)
{
if(S.top != -1)
return 0;
else
return 1;
}
//优先级函数
int Priority(char op)
{
switch(op)
{
case '(':
case '#':return 0;
case '+':
case '-':return 1;
case '*':
case '/':return 2;
}
return -1;
}
//中缀表达式 转 后缀表达式
void CTpostExp(SeqQueue *Q)
{
char c,t;
SeqStack S;
InitStack(&S);
do
{
c = getchar();
switch(c)
{
case '0':
case '1':
case '2':
case '3':
case '4':
case '5':
case '6':
case '7':
case '8':
case '9':EnQueue(Q,c); break;
case '(':Push(&S,c); break;
case ')':
do
{
t = Pop(&S);
EnQueue(Q,t);
}while(t != '(');
break;
case '+':
case '-':
case '*':
case '/':
if(S.top == -1) Push(&S,c);
else
{
while(Priority(c) <= Priority(GetTop(S)))
{
t = Pop(&S);
EnQueue(Q,t);
}
Push(&S,c);
}
break;
}
}while(c != '\n');
while(!EmptyStack(S))
{
t = Pop(&S);
EnQueue(Q,t);
}
}
//根据后缀表达式求值
void CaculatePostExp(SeqQueue Q)
{
char temp;
char num; //用于保存输入数据
int x,y,result;
SeqStack S;
InitStack(&S);
while(!QueueEmpty(Q))
{
temp = DeQueue(&Q);
switch(temp)
{
case '0':
case '1':
case '2':
case '3':
case '4':
case '5':
case '6':
case '7':
case '8':
case '9':Push(&S,temp); break;
case '+':
y = Pop(&S) - '0';
x = Pop(&S) - '0';
Push(&S,(char)(x + y + '0'));
break;
case '-':
y = Pop(&S) - '0';
x = Pop(&S) - '0';
Push(&S,(char)(x - y + '0'));
break;
case '*':
y = Pop(&S) - '0';
x = Pop(&S) - '0';
Push(&S,(char)(x * y + '0'));
break;
case '/':
y = Pop(&S) - '0';
x = Pop(&S) - '0';
Push(&S,(char)(x / y + '0'));
break;
}
}
temp = Pop(&S);
result = temp - '0';
printf("%d\n",result);
}
int main()
{
int n;
char temp;
SeqQueue Q;
scanf("%d%*c",&n);
while(n--)
{
InitQueue(&Q);
CTpostExp(&Q);
CaculatePostExp(Q);
}
}
4. Bracket matching, input the number of tests n, and then input n sets of samples in n lines. If they match, output yes, otherwise output no.
Input example:
2
([3])
([[[)
Output:
yes
no
Reference Code:
#include<stdio.h>
#include<string.h>
int main()
{
int n,i,len,flag,big,mid,small;
char a[100];
scanf("%d%*c",&n);
while(n--)
{
flag=1;
big=0;
mid=0;
small=0;
gets(a);
len=strlen(a);
for(i=0;i<len;i++)
{
switch(a[i])
{
case'{':
big++;
break;
case'[':
mid++;
break;
case'(':
small++;
break;
case'}':
if(big>0) big--;
else flag=0;
break;
case']':
if(mid>0) mid--;
else flag=0;
break;
case')':
if(small>0) small--;
else flag=0;
break;
}
}
if(big!=0||mid!=0||small!=0)
flag=0;
if(flag==1)
printf("yes\n");
if(flag==0)
printf("no\n");
}
}