输入一个百分制的成绩t，将其转换成对应的等级，C语言实现

Problem Description
输入一个百分制的成绩t，将其转换成对应的等级，具体转换规则如下：
90~100为A;
80~89为B;
70~79为C;
60~69为D;
0~59为E;

Input
输入数据有多组，每组占一行，由一个整数组成。

Output
对于每组输入数据，输出一行。如果输入数据不在0~100范围内，请输出一行：“Score is error!”。

Sample Input
56
67
100
123

Sample Output
E
D
A
Score is error!

https://www.cnblogs.com/wuyoucao/p/4562341.html

#include <stdio.h>
#include <math.h>
int main()
{
int n;
int x = 0;
int a[10] = { NULL };
scanf_s("%d", &n);
int i = 1;
while (i <= n)
{
scanf_s("%d", &a[x]);
i++;
if (a[x] <= 100 & a[x] >= 90)
printf("A\n");
else if (a[x] < 90 & a[x] >= 80)
printf("B\n");
else if (a[x] < 80 & a[x] >= 70)
printf("C\n");
else if (a[x] < 70 & a[x] >= 60)
printf("D\n");
else if (a[x] < 60 & a[x] >= 0)
printf("E\n");
else
printf("Score is error!\n");
}
return 0;
}

#include
#include

int main()
{
printf("Input(按任意字母键退出输入)\n");
int score[100];
int num=0;
while(scanf("%d",&score[num]))
{
num++;
}
printf("Output\n");
int i=0;
for(i=0;i if(score[i]100)
printf("Score is error!\n");
else if(score[i]<60)
printf("E\n");
else if(score[i]<70)
printf("D\n");
else if(score[i]<80)
printf("C\n");
else if(score[i]<90)
printf("B\n");
else if(score[i]<=100)
printf("A\n");
}
return 0;
}

#include<stdio.h>

int main()
{
printf("Input your score:\n");
int score[100],temp;
int num=0;
while(scanf("%d",&score[num]))
{
num++;
}
printf("Output\n");
int i=0;
for(i=0;i if(score[i]100)
{

temp=score[i]/10;
switch(score[i]):

case score (9 || 10):
printf("A\n");
continue;

case score (8):
printf("B\n");
continue;

case score(7):
printf("C\n");
continue;

//以此类推


}
return 0;
}

#include
#include
int main()
{
printf("Input(按任意字母键退出输入)\n");
int score[100];
int num=0;
while(scanf("%d",&score[num]))
{
num++;
}
printf("Output\n");
int i=0;
for(i=0;i if(score[i]100)
printf("Score is error!\n");
else if(score[i]<60)
printf("E\n");
else if(score[i]<70)
printf("D\n");
else if(score[i]<80)
printf("C\n");
else if(score[i]<90)
printf("B\n");
else if(score[i]<=100)
printf("A\n");
}
return 0;
}