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抄写例题作业
例题9.1
#include<stdio.h> int main() { struct Student { long int num; char name[20]; char sex; char addr[20]; }a={10101,"Li Lin",‘M‘,"123 Beijing Road"}; printf("NO.:%ld\nname:%s\nsex:%c\naddress:%s\n",a.num,a.name,a.sex,a.addr); return 0; }
NO.:10101
name:Li Lin
sex:M
address:123 Beijing Road
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Process exited after 0.02924 seconds with return value 0
请按任意键继续. . .
例题9.2
#include<stdio.h> int main() { struct Student { int num; char name[20]; float score; }student1,student2; scanf("%d%s%f",&student1.num,student1.name,&student1.score); scanf("%d%s%f",&student2.num,student2.name,&student2.score); printf("The higher score is:\n"); if(student1.score>student2.score) printf("%d %s %6.2f",student1.num,student1.name,student1.score); else if(student1.score<student2.score) printf("%d %s %6.2f",student2.num,student2.name,student2.score); else { printf("%d %s %6.2f",student1.num,student1.name,student1.score); printf("%d %s %6.2f",student2.num,student2.name,student2.score); } return 0; }
10101 Wang 89
10103 Ling 90
The higher score is:
10103 Ling 90.00
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Process exited after 12.16 seconds with return value 0
请按任意键继续. . .
例题9.3
#include<string.h> #include<stdio.h> struct Person { char name[20]; int count; }leader[3]={"Li",0,"Zhang",0,"Sun",0}; int main() { int i,j; char leader_name[20]; for(int i=1;i<=10;i++) { scanf("%s",leader_name); for(j=0;j<3;j++) if(strcmp(leader_name,leader[j].name)==0)leader[j].count++; } printf("\nResult\n"); for(i=0;i<3;i++) printf("%5s:%d\n",leader[i].name,leader[i].count); return 0; }
Li
Li
Sun
Zhang
Zhabg
Sun
Li
Sun
Zhang
Li
Result
Li:4
Zhang:2
Sun:3
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Process exited after 47.39 seconds with return value 0
请按任意键继续. . .
例题9.4
#include<stdio.h> struct Student { int num; char name[20]; float score; }; int main() { struct Student stu[5]={{10101,"Zhang",78},{10103,"Wang",98.5},{10106,"Li",86}, {10108,"Ling",73.5},{10110,"Sun",100}}; struct Student temp; const int n=5; int i,j,k; printf("The order is:\n"); for(i=0;i<n-1;i++) { k=i; for(j=i+1;j<n;j++) if(stu[j].score>stu[k].score) k=j; temp=stu[k];stu[k]=stu[i];stu[i]=temp; } for(i=0;i<n;i++) printf("%6d %8s %6.2f\n",stu[i].num,stu[i].name,stu[i].score); printf("\n"); return 0; }
The order is:
10110 Sun 100.00
10103 Wang 98.50
10106 Li 86.00
10101 Zhang 78.00
10108 Ling 73.50
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Process exited after 0.2451 seconds with return value 0
请按任意键继续. . .
例题9.5
#include<stdio.h> #include<string.h> int main() { struct Student { long num; char name[20]; char sex; float score; }; struct Student stu_1; struct Student *p; p=&stu_1; stu_1.num=10101; strcpy(stu_1.name,"Li Lin"); stu_1.sex=‘M‘; stu_1.score=89.5; printf("No.:%ld\nname:%s\nsex:%c\nscore:%5.1f\n", stu_1.num,stu_1.name,stu_1.sex,stu_1.score); printf("\nNo.:%ld\nname:%s\nsex:%c\nscore:%5.1f\n", (*p).num,(*p).name,(*p).sex,(*p).score); return 0; }
No.:10101
name:Li Lin
sex:M
score: 89.5
No.:10101
name:Li Lin
sex:M
score: 89.5
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Process exited after 0.1452 seconds with return value 0
请按任意键继续. . .
抄写例题作业