int[] stuGrad = new int[]{58,59,60,61,62}; int moreThan = 0 int lessThan for(int i = 0;i<stuGrad.Length;i++) { if(0<strGrad[i]<60) { lessThan++; } else if(strGrad[i]<70) { moreThan++ } } Console.WriteLine("0-50:{0}",lessThan); Console.WriteLine("60-69:{0}",moreThan);
Int32[] counters = new Int32[5]; String str; Console.WriteLine("输入分数(-1结束)"); while (true) { str = Console.ReadLine(); Int32 i = Convert.ToInt32(str); if (i == -1) break; Int32 ci = i / 10; if (ci == 10) ci--; if (ci <= 5) counters[0]++; else counters[ci - 5]++;
如果是关系数据库,为什么不用SQL语句呢,这样速度更快些比如查 select count(*) as 分数0到59 from table where 分数>0 and 分数<=59 ,以此类推,这样几条语句全部搞定。
方案一:类似5楼(LessThanJiGe;MoreThanJiGe) 方案二:用SQL实现 select count(*) as Result from (select (case when score<=59 then 0 else 1 end) as JiGe) from M_Student) a group by a.JiGe 注:0代表不及格 ,1代表及格 结果Example: Result 32 168
int[] stuGrad;
stuGrad=new int[100];for(int i=0;i<stuGrad.Length;i++)
{
if()
}
int[] stuGrad = new int[]{58,59,60,61,62};
int moreThan = 0
int lessThan
for(int i = 0;i<stuGrad.Length;i++)
{
if(0<strGrad[i]<60)
{
lessThan++;
}
else if(strGrad[i]<70)
{
moreThan++
}
}
Console.WriteLine("0-50:{0}",lessThan);
Console.WriteLine("60-69:{0}",moreThan);
Console.WriteLine("输入分数(-1结束)");
while (true)
{
str = Console.ReadLine();
Int32 i = Convert.ToInt32(str);
if (i == -1)
break;
Int32 ci = i / 10;
if (ci == 10)
ci--; if (ci <= 5)
counters[0]++;
else
counters[ci - 5]++;
} Console.WriteLine("0-59 : {0}人",counters[0]);
Console.WriteLine("60-69 : {0}人", counters[1]);
Console.WriteLine("70-79 : {0}人", counters[2]);
Console.WriteLine("80-89 : {0}人", counters[3]);
Console.WriteLine("90-100 : {0}人", counters[4]);
方案二:用SQL实现
select count(*) as Result from (select (case when score<=59 then 0 else 1 end) as JiGe) from M_Student) a group by a.JiGe
注:0代表不及格 ,1代表及格
结果Example:
Result
32
168