# include≤ stdio.h> int maino d struct Student dint num; char name[20] float scorer student 1 studenti scanf(%/od%/os%/of &student1num student1. name, &student1 score); sc! nf(o/od%/os%/f", &student2num, student name,&student2score); 不能加&
#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); 不能加&
printf( the higher score is: n i if (student1 score>student2score) printf( %/od %/os%/06.2f\n" student1num student1. name, student1 score); else if(student1 score<student2 score) printf( %od %/os %/06.2f\n"student2,num student2.name, student2 score); else printf(%/od %os%/6.2f(n"student1num, student1.name, student1 score); printf( o/od %/os%/6.2f\n"student2 num student2. name, student2 score)i 10101 Wang 89 return o: 10103 Ling 90 he higher score is: 10103 Ling 90.00
printf("The higher score is:\n"); if (student1.score>student2.score) printf("%d %s %6.2f\n",student1.num, student1.name, student1.score); else if (student1.score<student2.score) printf("%d %s %6.2f\n",student2.num, student2.name, student2.score); else {printf("%d %s %6.2f\n",student1.num, student1.name, student1.score); printf("%d %s %6.2f\n",student2.num, student2.name, student2.score); } return 0; }
92纬构体数组 例93有3个候选人,每个选民只能投票 选一人,要求编一个统计选票的程序,先 后输入被选人的名字,最后输出各人得票 结果
9.2 结构体数组 例9.3 有3个候选人，每个选民只能投票 选一人，要求编一个统计选票的程序，先 后输入被选人的名字，最后输出各人得票 结果
include <string. h> # include≤ stdio. h> struct person char name[20] int count: Leader[3]=t li o, zhang,o Sun",OSi 全局的结构体数组 name count leader[o] Zhane Sun 000
#include <string.h> #include <stdio.h> struct Person { char name[20]; int count; }leader[3]={“Li”,0,“Zhang”,0,“Sun”,0}; 全局的结构体数组 name count leader[0] Li 0 Zhang 0 Sun 0
int maino d int iii char leader name[20]; for(i=1;<=10++) d scanf(%/os" leader name)i for(j=0厅<3厅j++) if(strcmp(leader name, leader[jl. name)==0) leader[j]。 count++; for(i=0;i<3i++) printf(%/o5s: %/od\n leader[il name r eader[i]。 count); return os
int main() { int i,j; char leader_name[20]; for (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++; } for(i=0;i<3;i++) printf("%5s:%d\n“,leader[i].name, leader[i].count); return 0; }