1.打分统计并按格式输出
for (i = 0; i < M; i++)
{
switch (feedback[i])
{
case 1: count[1]++; break;
case 2: count[2]++; break;
case 3: count[3]++; break;
case 4: count[4]++; break;
case 5: count[5]++; break;
case 6: count[6]++; break;
case 7: count[7]++; break;
case 8: count[8]++; break;
case 9: count[9]++; break;
case 10: count[10]++; break;
default:printf("input error! \n");
}
}
还可以用一行解决:
for (i = 0; i < M; i++)
{
count[feedback[i]]++;
}
紧接着可以找出这些数的众数(Mode)
int max = 0;
for (grade = 1; grade <= N-1; grade++)
{
if (count[grade] > max)
{
max = count[grade];
modeValue = grade; //下标
}
}
return modeValue;
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