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这里写目录标题
题目链接
- 平均平均⾝⾼00:00:00⸺00:06:05题号:(⽆)链接:https://www.nowcoder.com/questio
nTerminal/487e757828ed4d40aa391b305105df68 - HTTP状态码00:06:05⸺00:13:38题号:BC69链接:https://www.nowcoder.com/practic
e/99dba043761e43c2a6f931e2c5c247c7?tpId=290&tqId=39857&ru=/exam/oj - 数字三⻆形00:13:38⸺00:18:15题号:BC113链接:https://www.nowcoder.com/practice/804a22929b844e6b9379a5e90b5e2197?
- 公务员⾯试00:18:15⸺00:26:15题号:BC93链接:https://www.nowcoder.com/practic
e/f3a134908d5b41869f14f58307008a97?tpId=290&tqId=39881&ru=/exam/oj - 有序序列插⼊⼀个数00:26:15⸺00:43:25题号:BC123链接:https://www.nowcoder.co
m/practice/444e87f938464906a1649cff236b102b?tpId=290&tqId=39911&ru=/exam/oj - 筛选法求素数00:43:25⸺01:07:19题号:BC127链接:https://www.nowcoder.com/pra
ctice/06c3dcc303654ef4926654023eca1e5a?tpId=290&tqId=39915&ru=/exam/oj - 图像相似度01:07:19⸺01:14:20题号:BC135链接:https://www.nowcoder.com/practic
e/f2952ee3bb5c48a9be6c261e29dd1092?tpId=290&tqId=39923&ru=/exam/oj - 登录验证01:14:20⸺01:21:50题号:BC144链接:https://www.nowcoder.com/practic
e/799daf76369c4fb1a2e2d56f885225ae?tpId=290&tqId=39932&ru=/exam/oj - 包含数字9的数01:21:50⸺01:27:20题号:BC89链接:https://www.nowcoder.com/pra
ctice/0948f4f3344c452f843afd3585dd0f8d?tpId=290&tqId=39877&ru=/exam/oj - 奇偶统计01:27:20⸺01:34:39题号:BC80链接:https://www.nowcoder.com/practic
e/04de8eb0ecab426fa6be3ae99af17210?tpId=290&tqId=39868&ru=/exam/oj
T1:
解:
c
#include <stdio.h>
int main() {
double arr[5] = {0};
int i = 0;
double sum = 0;
for(i = 0;i<5;i++)
{
scanf("%lf",&arr[i]);
sum+=arr[i];
}
printf("%.2lf\n",sum/5.0);
return 0;
}
一般可能会搞个数组存一下数据然后进行处理,其实这个数据没必要存储,节约内存空间,提升效率,可以修改一下代码:
c
#include <stdio.h>
int main() {
int i = 0;
double sum = 0;
double m = 0.0;
for(i = 0;i<5;i++)
{
scanf("%lf",&m);
sum+=m;
}
printf("%.2lf\n",sum/5.0);
return 0;
}
T2:
解:
c
#include <stdio.h>
int main()
{
int n;
while (scanf("%d", &n) != EOF)
{
switch(n)
{
case 200:
printf("OK\n");
break;
case 202:
printf("Accepted\n");
break;
case 400:
printf("Bad Request\n");
break;
case 403:
printf("Forbidden\n");
break;
case 404:
printf("Not Found\n");
break;
case 500:
printf("Internal Server Error\n");
break;
case 502:
printf("Bad Gateway\n");
break;
}
}
return 0;
}
T3:
解:
c
#include <stdio.h>
int main()
{
int n = 0;
while(~scanf("%d",&n))
{
int i = 0;
int j = 0;
for(i=0;i<n;i++)
{
for(j=1;j<=i+1;j++)
{
printf("%d ",j);
}
printf("\n");
}
}
return 0;
}
T4:
c
#include <stdio.h>
int main() {
int arr[7] = {0};
while (scanf("%d %d %d %d %d %d %d", &arr[0], &arr[1], &arr[2], &arr[3],
&arr[4], &arr[5], &arr[6]) != EOF) {
int i = 0;
int max = 0;
int min = 100;
double sum = 0.0;
for (i = 0; i < 7; i++) {
if (arr[i] > max)
max = arr[i];
if (arr[i] < min)
min = arr[i];
sum += arr[i];
}
sum -= (max + min);
printf("%.2lf\n", sum / 5.0);
}
return 0;
}
T5:
解:
c
#include <stdio.h>
int main()
{
int arr[51] = {0};
int i = 0;
int n = 0;
while(scanf("%d",&n)!=EOF)
{
//输入数据
for(i = 0;i<n;i++)
{
scanf("%d",&arr[i]);
}
//插入数据
int q = 0;
scanf("%d",&q);
for(i = 0;i<n;i++)
{
if(q<arr[i])
{
int j = 0;
//往后挪动一位
for(j=n;j> i;j--)
{
arr[j] = arr[j-1];
}
arr[i] = q;
break;
}
else
{
;
}
}
//特殊情况
if(i==n)
{
arr[i] = q;
}
for(i = 0;i<n+1;i++)
{
printf("%d ",arr[i]);
}
}
return 0;
}
T6:
0%任何非0数字=0!
//一般这个筛选法比较麻烦
c
#include <stdio.h>
int main()
{
int n = 0;
int arr[101] = {0};
int count = 0;
while(~scanf("%d",&n))
{
//存入数据
int i = 0;
for(i = 2;i<=n;i++)
{
arr[i] = i;
}
//生成除数,顺便置换为0
int j = 0;
for(j = 2;j< n;j++)
{
for(i = j+1;i<= n;i++)
{
if(arr[i]&&arr[i]%j==0)
{
arr[i] = 0;
count++;
}
}
}
for(i = 0;i<=n;i++)
{
if(arr[i])
{
printf("%d ",arr[i]);
}
}
printf("\n%d",count);
}
return 0;
}
//试除法比较简单:
T7:
c
#include <stdio.h>
int main() {
int m = 0;
int n = 0;
while(scanf("%d %d",&m,&n)!=EOF)
{
int count = 0;
int arr1[m][n];
int arr2[m][n];
int i = 0;
int j = 0;
for(i = 0;i<m;i++)
{
for(j = 0;j<n;j++)
{
scanf("%d",&arr1[i][j]);
}
}
for(i = 0;i<m;i++)
{
for(j = 0;j<n;j++)
{
scanf("%d",&arr2[i][j]);
}
}
for(i = 0;i<m;i++)
{
for(j = 0;j<n;j++)
{
if(arr1[i][j]==arr2[i][j])
{
count++;
}
}
}
printf("%.2lf",100.0*count/(m*n));
}
return 0;
}
T8:
解:
c
#include <stdio.h>
#include<string.h>
int main()
{
char a[100] = {0};
char b[100] = {0};
while (scanf("%s %s", &a[0], &b[0]) != EOF) { // 注意 while 处理多个 case
// 64 位输出请用 printf("%lld") to
if(!strcmp(a,"admin")&&!strcmp(b,"admin"))
{
printf("Login Success!\n");
}
else {
{
printf("Login Fail!\n");
}
}
}
return 0;
}
T9:
解:
c
#include <stdio.h>
int main()
{
int count = 0;
int i = 0;
for(i = 1;i<=2019;i++)
{
int t = i;
while(t)
{
if(t%10==9)
{
count++;
break;
}
else
{
t/=10;
}
}
}
printf("%d",count);
return 0;
}
T10:
c
#include <stdio.h>
int main() {
int n = 0;
int i = 0;
scanf("%d",&n);
int odd= 0;
for(i = 0;i<=n;i++)
{
if(i%2)
{
odd++;
}
}
printf("%d %d",odd,n-odd);
return 0;
}
改进:
c
#include <stdio.h>
int main() {
int n = 0;
int i = 0;
scanf("%d",&n);
int odd= 0;
if(n%2)
{
odd = n/2+1;
}
else {
odd=n/2;
}
printf("%d %d",odd,n-odd);
return 0;
}