解题思路:
注意事项:
参考代码:
#include<stdio.h>
struct garde {
int sum;
int xuehao;
};
int main(){
int arr[1000] = { 0 };
int brr[1000] = { 0 };
int crr[1000] = { 0 };
int n;
int temp1;
struct garde su[1000];
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
scanf("%d", &arr[i]);
scanf("%d", &brr[i]);
scanf("%d", &crr[i]);
}
for (int i = 1; i <= n; i++) {
su[i].sum = arr[i] + brr[i] + crr[i];
su[i].xuehao = i;
}
for (int i = 1; i <= n; i++) {
for (int k = 1; k <= n - 1; k++) {
if (su[k].sum < su[k + 1].sum) {
temp1 = su[k].sum;
su[k].sum = su[k + 1].sum;
su[k + 1].sum = temp1;
temp1 = su[k].xuehao;
su[k].xuehao = su[k + 1].xuehao;
su[k + 1].xuehao = temp1;
temp1 = arr[k];
arr[k] = arr[k + 1];
arr[k+1] = temp1;
}
}
}
for (int k = 1; k <= n - 1; k++) {
if (su[k].sum == su[k + 1].sum&&arr[k]<arr[k+1]) { //对总分相同的两个学生根据语文成绩排序
temp1 = su[k].sum;
su[k].sum = su[k + 1].sum;
su[k + 1].sum = temp1;
temp1 = su[k].xuehao;
su[k].xuehao = su[k + 1].xuehao;
su[k + 1].xuehao = temp1;
temp1 = arr[k];
arr[k] = arr[k + 1];
arr[k + 1] = temp1;
}
}
if (n >= 5) { //判断学生总数是否大于5(所给的样例可能都是大于五的,这步可能不需要)
for (int i = 1; i <= 5; i++) {
printf("%d ", su[i].xuehao); //根据格式进行输出
printf("%d\n", su[i].sum);
}
}
else if (n < 5) {
for (int i = 1; i <= n; i++) {
printf("%d ", su[i].xuehao);
printf("%d\n", su[i].sum);
}
}
return 0;
}
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