6-4 电码加密 (10分)
为了防止信息被别人轻易窃取，需要把电码明文通过加密方式变换成为密文。要求编写并调用函数encrypt ()，按要求修改字符串内容。变换规则如下：小写字母z变换成为a，其他字母变换成为该字母ASCII码顺序后一位的字母，比如o变换成为p。
函数接口定义：
void encrypt ( char *s);
其中 s为字符串首地址。
裁判测试程序样例：
#include <stdio.h>
# include <string.h>
# define MAXLINE 80
void encrypt ( char *s);
int main (void)
{
char line [MAXLINE];
gets(line);
encrypt (line);
printf ("%s\n", line);
return 0;
}
/* 请在这里填写答案 */
输入样例：
adfz
输出样例：
bega
答案代码：
void encrypt ( char *s)
{
int n = strlen(s);
int i;
for(i = 0;i < n;i++)
{
if(s[i] == 'z')
{
s[i] = 'a';
}
else
{
s[i] = s[i] + 1;
}
}
}
6-4 输入年份和天数，输出对应的年、月、日 (15分)
函数接口定义：
void month_day ( year, yeardy, *pmonth, *pday);
其中year是年，yearday是天数，pmonth和pday是计算得出的月和日。
裁判测试程序样例：
#include <stdio.h>
void month_day ( int year, int yearday, int * pmonth, int * pday);
int main (void)
{
int day, month, year, yearday; /* 定义代表日、月、年和天数的变量*/
scanf ("%d%d", &year, &yearday );
month_day (year, yearday, &month, &day );/* 调用计算月、日函数 */
printf ("%d %d %d\n", year, month, day );
return 0;
}
/* 请在这里填写答案 */
输入样例：
2000 61
输出样例：
2000 3 1
答案代码：
void month_day ( int year, int yearday, int * pmonth, int * pday)
{
if((year%4 == 0 && year%100 != 0) || year%400 == 0)
{
if(yearday <= 31)
{
*pmonth = 1;
*pday = yearday;
}
else if(yearday > 31 && yearday <= 60)
{
*pmonth = 2;
*pday = yearday - 31;
}
else if(yearday > 60 && yearday <= 91)
{
*pmonth = 3;
*pday = yearday - 60;
}
else if(yearday > 91 && yearday <= 121)
{
*pmonth = 4;
*pday = yearday - 91;
}
else if(yearday > 121 && yearday <= 152)
{
*pmonth = 5;
*pday = yearday - 121;
}
else if(yearday > 152 && yearday <= 182)
{
*pmonth = 6;
*pday = yearday - 152;
}
else if(yearday > 182 && yearday <= 213)
{
*pmonth = 7;
*pday = yearday - 182;
}
else if(yearday > 213 && yearday <= 244)
{
*pmonth = 8;
*pday = yearday - 213;
}
else if(yearday > 244 && yearday <= 274)
{
*pmonth = 9;
*pday = yearday - 244;
}
else if(yearday > 274 && yearday <= 305)
{
*pmonth = 10;
*pday = yearday - 274;
}
else if(yearday > 305 && yearday <= 335)
{
*pmonth = 11;
*pday = yearday - 305;
}
else if(yearday > 335 && yearday <= 366)
{
*pmonth = 12;
*pday = yearday - 335;
}
}
else
{
if(yearday <= 31)
{
*pmonth = 1;
*pday = yearday;
}
else if(yearday > 31 && yearday <= 59)
{
*pmonth = 2;
*pday = yearday - 31;
}
else if(yearday > 59 && yearday <= 90)
{
*pmonth = 3;
*pday = yearday - 59;
}
else if(yearday > 90 && yearday <= 120)
{
*pmonth = 4;
*pday = yearday - 90;
}
else if(yearday > 120 && yearday <= 151)
{
*pmonth = 5;
*pday = yearday - 120;
}
else if(yearday > 151 && yearday <= 181)
{
*pmonth = 6;
*pday = yearday - 151;
}
else if(yearday > 181 && yearday <= 212)
{
*pmonth = 7;
*pday = yearday - 181;
}
else if(yearday > 212 && yearday <= 243)
{
*pmonth = 8;
*pday = yearday - 212;
}
else if(yearday > 243 && yearday <= 273)
{
*pmonth = 9;
*pday = yearday - 243;
}
else if(yearday > 273 && yearday <= 304)
{
*pmonth = 10;
*pday = yearday - 273;
}
else if(yearday > 304 && yearday <= 334)
{
*pmonth = 11;
*pday = yearday - 304;
}
else if(yearday > 334 && yearday <= 365)
{
*pmonth = 12;
*pday = yearday - 334;
}
}
}
6-4 二分查找 (15分)
已有一个10个元素的整形数组a，且按值从小到大有序。输入一个整数x，然后在数组中查找x，如果找到，输出相应的下标，否则，输出"Not Found"。。 要求编写函数int Bsearch(int *p, int n, int x)，找到返回下标，找不到返回-1。
函数接口定义：
int Bsearch(int *p, int n, int x);
其中 p是数组首地址，n是数组元素个数，x是要查找的值。找到返回下标，找不到返回-1。
裁判测试程序样例：
#include<stdio.h>
int Bsearch(int *p, int n, int x);
int main(void)
{
int a[10] = {1,2,3,4,5,6,7,8,9,10};
int x, m;
scanf("%d",&x);
m = Bsearch(a, 10, x);
if(m >= 0)
printf("Index is %d\n",m);
else
printf( "Not Found\n");
return 0;
}
/* 请在这里填写答案 */
输入样例：
8
输出样例：
Index is 7
答案代码：
int Bsearch(int *p, int n, int x)
{
int high = n-1;
int low = 0;
int mid = (low+high)/2;
if(p[high] < x)
return -1;
if(p[low] > x)
return -1;
while(p[mid] != x)
{
if(p[mid] > x)
{
high = mid;
}
else
{
low = mid;
}
mid = (low+high)/2;
if(high == low)
return -1;
}
return mid;
}
6-24 使用函数的选择法排序 (25分)
本题要求实现一个用选择法对整数数组进行简单排序的函数。
函数接口定义：
void sort( int a[], int n );
其中a是待排序的数组，n是数组a中元素的个数。该函数用选择法将数组a中的元素按升序排列，结果仍然在数组a中。
裁判测试程序样例：
#include <stdio.h>
#define MAXN 10
void sort( int a[], int n );
int main()
{
int i, n;
int a[MAXN];
scanf("%d", &n);
for( i=0; i<n; i++ )
scanf("%d", &a[i]);
sort(a, n);
printf("After sorted the array is:");
for( i = 0; i < n; i++ )
printf(" %d", a[i]);
printf("\n");
return 0;
}
/* 你的代码将被嵌在这里 */
4
5 1 7 6
输出样例：
After sorted the array is: 1 5 6 7
答案代码：
void sort( int a[], int n )
{
int i;
int j;
int flag;
for(i = 0;i < n;i++)
{
flag = i;
for(j = i;j < n;j++)
{
if(a[j] < a[flag])
{
flag = j;
}
}
int t = a[i];
a[i] = a[flag];
a[flag] = t;
}
}
