如何统计一列中某个数据连续出现N次的情况出现的次数
如何统计一列中某个数据连续出现N次的情况出现的次数
int num[100][100]={0};//你可自己定义数组大小,理论上是没有上限的,我试
了下,10000多每问题
//输入你自己写
//0次,1次......10次以上
int
times0,times1,times2,times3,times4,times5,times6,times7,times8,times9,
times10,timesOver10;
//初始化
times0=times1=times2=times3=times4=times5=times6=times7=times8=times9=
times10=timesOver10=0;
for(int i=0;i<100;i++)
{
int timesTemp=0;
for(int j=0;j<100;j++)
{
if(j==1&&num[i][j]==0)
timesTemp++;
else
{
if(j!=0&&num[i][j]==0)
{
if(num[i][j-1]==0)
timesTemp++;
}
else timesTemp=0;
}
}
switch(timesTemp)
{
case 0:times0++;break;
case 1:times1++;break;
case 2:times2++;break;
case 3:times3++;break;
case 4:times4++;break;
case 5:times5++;break;
case 6:times6++;break;
case 7:times7++;break;
case 8:times8++;break;
case 9:times9++;break;
case 10:times10++;break;
default:timesOver10++;
}
}
//如果你用C++,就用以下方式输出:
cout<<"0次:"<<times0<<endl;
cout<<"1次:"<<times1<<endl;
cout<<"2次:"<<times2<<endl;
cout<<"3次:"<<times3<<endl;
cout<<"4次:"<<times4<<endl;
cout<<"5次:"<<times5<<endl;
cout<<"6次:"<<times6<<endl;
cout<<"7次:"<<times7<<endl;
cout<<"8次:"<<times8<<endl;
cout<<"9次:"<<times9<<endl;
cout<<"10次:"<<times10<<endl;
cout<<"10次以上:"<<timesOver10<<endl;
//如果你用C,就用以下方式输出:
printf("0次:%d
",times0);
printf("1次:%d
",times1);
printf("2次:%d
",times2);
printf("3次:%d
",times3);
printf("4次:%d
",times4);
printf("5次:%d
",times5);
printf("6次:%d
",times6);
printf("7次:%d
",times7);
printf("8次:%d
",times8);
printf("9次:%d
",times9);
printf("10次:%d
",times10);
printf("10次以上:%d
",timesOver10);
比《请问电子表格列中数字连续出现了1次的总次数,2次的总次数,3次的总次数以此类推的公式》又深了。