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给定一个十进制正整数N,写下从1开始,到N的所有正整数,然后数一下其中出现的所有“1”的个数。例如:N =12,我们会写下1,2,3,4,5,6,7,8,9,10,11,12。这样1出现的个数为5
java代码实现
public class demo1 {
public static void main(String[] args) {
Sum sum = new Sum();
System.out.println("请输入一个正整数N:");
int v = new Scanner(System.in).nextInt();
System.out.println("从1开始,到N的所有整数,1出现的个数:" + sum.sum1(v));
}
}
class Sum{
public int sum1(int n){
int count = 0;
int factor =1;
int lowerNum = 0;
int currNum = 0;
int higherNum = 0;
while(n/factor !=0){
lowerNum = n-(n/factor)*factor;
currNum = (n/factor)%10;
higherNum = n/(factor*10);
switch(currNum){
case 0:
count +=higherNum*factor;
break;
case 1:
count +=higherNum*factor+lowerNum+1;
break;
default:
count +=(higherNum+1)*factor;
break;
}
factor *=10;
}
return count;
}
}
运行结果:请输入一个正整数N:
256
从1开始,到N的所有整数,1出现的个数:156