抄写例题作业1
Posted Dong诗原
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1.例9.1
（1）代码实现
1 #include<stdio.h> 2 int main() 3 { 4 struct stu 5 { 6 long int num; 7 char name[20]; 8 char sex[3]; 9 char addr[20]; 10 }a={1010,"董诗原","男","hkj"}; 11 printf("No:%d\nname:%s\nsex:%s\naddr:%s",a.num,a.name,a.sex,a.addr); 12 }
（2）运行结果
No:1010 name:董诗原 sex:男 addr:hkj -------------------------------- Process exited after 0.4101 seconds with return value 0 请按任意键继续. . .
2.例9.2
（1）代码实现
1 #include<stdio.h> 2 int main() 3 { 4 struct stu 5 { 6 long int num; 7 char name[20]; 8 float score; 9 }stu1,stu2; 10 scanf("%d%s%f",&stu1.num,stu1.name,&stu1.score); 11 scanf("%d%s%f",&stu2.num,stu2.name,&stu2.score); 12 printf("The higher score is:\n"); 13 if(stu1.score>stu2.score) 14 printf("%d %s %.2f\n",stu1.num,stu1.name,stu1.score); 15 else if(stu1.score<stu2.score) 16 printf("%d %s %.2f\n",stu2.num,stu2.name,stu2.score); 17 else{ 18 printf("%d %s %.2f\n",stu1.num,stu1.name,stu1.score); 19 printf("%d %s %.2f\n",stu2.num,stu2.name,stu2.score); 20 } 21 }
（2）运行结果
10101 wang 89 10103 ling 90 The higher score is: 10103 ling 90.00 -------------------------------- Process exited after 25.17 seconds with return value 0 请按任意键继续. . .
3.例9.3
（1）代码实现
1 #include<stdio.h> 2 #include<string.h> 3 int main() 4 { 5 struct per{ 6 char name[20]; 7 int count; 8 }a[3]={"li",0,"zhang",0,"sun",0}; 9 char b[20]; 10 for(int i=0;i<10;i++) 11 { 12 scanf("%s",&b); 13 if(strcmp(b,a[0].name)==0) 14 a[0].count++; 15 else if(strcmp(b,a[1].name)==0) 16 a[1].count++; 17 else 18 a[2].count++; 19 } 20 printf("\nResult:\nli:%d\nzhang:%d\nsun:%d\n",a[0].count,a[1].count,a[2].count); 21 }
（2）运行结果
li li sun zhang zhang sun li sun zhang li Result: li:4 zhang:3 sun:3 -------------------------------- Process exited after 55.91 seconds with return value 0 请按任意键继续. . .
4.例9.4
（1）代码实现
1 #include<stdio.h> 2 struct stu 3 { 4 long int num; 5 char name[20]; 6 float score; 7 }; 8 int main() 9 { 10 struct stu t; 11 struct stu s[5]={10101,"zhang",78,10103,"wang",98.5,10106,"li",86,10108,"ling",73.5,10100,"sun",100 }; 12 for(int i=0;i<5;i++){ 13 for(int j=1;j<5-i;j++) 14 { 15 if(s[j].score>s[j-1].score) 16 { 17 t=s[j]; 18 s[j]=s[j-1]; 19 s[j-1]=t; 20 } 21 } 22 } 23 for(int i=0;i<5;i++) 24 { 25 printf("%d%8s%8.2f\n",s[i].num,s[i].name,s[i].score); 26 } 27 }
（2）运行结果
10100 sun 100.00 10103 wang 98.50 10106 li 86.00 10101 zhang 78.00 10108 ling 73.50 -------------------------------- Process exited after 0.46 seconds with return value 0 请按任意键继续. . .
5.例9.5
（1）代码实现
1 #include<stdio.h> 2 #include<string.h> 3 int main() 4 { 5 struct stu 6 { 7 long int num; 8 char name[20]; 9 char sex; 10 float score; 11 }; 12 struct stu stu_1; 13 struct stu *p; 14 p=&stu_1; 15 strcpy(stu_1.name,"Li lin"); 16 stu_1.num=10101; 17 stu_1.sex=‘M‘; 18 stu_1.score=89.5; 19 printf("No:%d\nname:%s\nsex:%c\nscore:%.2f\n\n",stu_1.num,stu_1.name,stu_1.sex,stu_1.score); 20 printf("No:%d\nname:%s\nsex:%c\nscore:%.2f",(*p).num,(*p).name,(*p).sex,(*p).score); 21 }
（2）运行结果
No:10101 name:Li lin sex:M score:89.50 No:10101 name:Li lin sex:M score:89.50 -------------------------------- Process exited after 0.4481 seconds with return value 0 请按任意键继续. . .
6.例9.6
（1）代码实现
1 #include<stdio.h> 2 #include<string.h> 3 struct stu 4 { 5 long int num; 6 char name[20]; 7 char sex; 8 int age; 9 }; 10 struct stu a[3]{10101,"Li lin",‘M‘,18,10102,"Zhang fang",‘M‘,19,10104,"Wang min",‘F‘,20}; 11 int main() 12 { 13 struct stu *p; 14 printf(" No. name sex age\n"); 15 for(p=a;p<a+3;p++) 16 printf("%5d%13s%5c%7d\n",p->num,p->name,p->sex,p->age); 17 }
（2）运行结果
No. name sex age 10101 Li lin M 18 10102 Zhang fang M 19 10104 Wang min F 20 -------------------------------- Process exited after 0.1007 seconds with return value 0 请按任意键继续. . .
7.例9.7
（1）代码实现
1 #include<stdio.h> 2 #include<string.h> 3 struct stu 4 { 5 long int num; 6 char name[20]; 7 float score[3]; 8 float aver; 9 }; 10 struct stu a[3]; 11 int main() 12 { 13 printf("请输入各学生信息：学号，姓名，三门课成绩：\n"); 14 for(int i=0;i<3;i++) 15 { 16 a[i].aver=0; 17 scanf("%d%s",&a[i].num,a[i].name); 18 for(int j=0;j<3;j++) 19 { 20 scanf("%f",&a[i].score[j]); 21 a[i].aver=a[i].aver+a[i].score[j]; 22 } 23 a[i].aver/=3; 24 } 25 int max=a[0].aver; 26 int count; 27 for(int i=1;i<3;i++) 28 { 29 if(a[i].aver>max){ 30 max=a[i].aver; 31 count=i; 32 } 33 } 34 printf("\n成绩最高的学生是：\n"); 35 printf("学号：%d\n姓名：%s\n",a[count].num,a[count].name); 36 printf("三门课成绩: "); 37 for(int i=0;i<3;i++) 38 printf("%.1f ",a[count].score[i]); 39 printf("\n平均成绩：%.2f",a[count].aver); 40 }
（2）运行结果
请输入各学生信息：学号，姓名，三门课成绩： 10101 li 78 89 98 10103 wang 98.5 87 69 10106 sun 88 76.5 89 成绩最高的学生是： 学号：10101 姓名：li 三门课成绩: 78.0 89.0 98.0 平均成绩：88.33 -------------------------------- Process exited after 64.47 seconds with return value 0 请按任意键继续. . .
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