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#2
遗情处有诗章2018-03-27 19:12
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import java.util.Scanner;
public class Test {
public static void main(String[] args){
Scanner in=new Scanner(System.in);
int n=in.nextInt();
int d=in.nextInt();
System.out.print(countDigit(n, d));
}
public static int countDigit( int n, int d)
{
Scanner in=new Scanner(System.in);
int count=in.nextInt();
int temp=in.nextInt();
int x=in.nextInt();
x=n;
if(x<0)
x=-x;
do{
temp=x%10;
if(temp==d)
count++;
x=x/10;
}
while(x>0);
return count;
}
}
public class Test {
public static void main(String[] args){
Scanner in=new Scanner(System.in);
int n=in.nextInt();
int d=in.nextInt();
System.out.print(countDigit(n, d));
}
public static int countDigit( int n, int d)
{
Scanner in=new Scanner(System.in);
int count=in.nextInt();
int temp=in.nextInt();
int x=in.nextInt();
x=n;
if(x<0)
x=-x;
do{
temp=x%10;
if(temp==d)
count++;
x=x/10;
}
while(x>0);
return count;
}
}
给定两个均不超过9的正整数a和n,要求编写函数求a+aa+aaa++⋯+aa⋯a(n个a)之和。函数接口定义:int fn( int a, int n ); int SumA( int a, int n );
其中函数fn须返回的是n个a组成的数字,比如,fn(2,3)返回的是222;
SumA返回要求的和,比如sumA(2,3)返回2 + 22 + 222。

import java.util.Scanner;
public class Test {
static int fn( int a, int n ) {
Scanner in=new Scanner(System.in);
int sum = in.nextInt();
sum=0;
int b = in.nextInt();
b=1;
for(int i=0;i<n;i++)
{
//b=b*10;
sum=sum+a*b;
b=b*10;
}
return sum;
}
static int SumA( int a, int n ) {
Scanner in=new Scanner(System.in);
int sum1 = in.nextInt();
sum1=0;
int sum2 = in.nextInt();
for(int i=1;i<n+1;i++)
{
sum2=fn(a,i);
sum1=sum1+sum2;
}
return sum1;
public static void main(String[] args){
Scanner in=new Scanner(System.in);
int a = in.nextInt();
int n = in.nextInt();
System.out.println(fn(a,n));
System.out.println(SumA(a,n));
}
}
public class Test {
static int fn( int a, int n ) {
Scanner in=new Scanner(System.in);
int sum = in.nextInt();
sum=0;
int b = in.nextInt();
b=1;
for(int i=0;i<n;i++)
{
//b=b*10;
sum=sum+a*b;
b=b*10;
}
return sum;
}
static int SumA( int a, int n ) {
Scanner in=new Scanner(System.in);
int sum1 = in.nextInt();
sum1=0;
int sum2 = in.nextInt();
for(int i=1;i<n+1;i++)
{
sum2=fn(a,i);
sum1=sum1+sum2;
}
return sum1;
public static void main(String[] args){
Scanner in=new Scanner(System.in);
int a = in.nextInt();
int n = in.nextInt();
System.out.println(fn(a,n));
System.out.println(SumA(a,n));
}
}