每日一题
题目
设N是一个四位数,它的9倍恰好是其反序数(例如:1234 的反序数是4321),
求N的值。
Python解题
for i in range(1000,100000):
base = i
res = i * 9
a = base // 1000 == res % 10
b = base % 1000 // 100 == res % 100 //10
c = base % 100 //10 == res % 1000 // 100
d = base % 10 == res // 1000
if a and b and c and d :
print(base)
JavaScript解题
<script>
for(var i=1000;i<10000;i++){
var ii = i * 9;
var a = Math.floor(i/1000) == ii % 10;
var b = Math.floor(i%1000/100) == Math.floor(ii%100/10);
var c = Math.floor(i%100/10) == Math.floor(ii%1000/100);
var d = i % 10 == Math.floor(ii/1000);
if(a&&b&&c&&d){
console.log(i)
}
}
</script>
Java解题
package code01;
public class Test03 {
public static void main(String[] args) {
boolean flag = true;
for(int i=1000;i<10000;i++) {
int ii = i * 9;
int[] base = {i/1000,i%1000/100,i%100/10,i%10};
int[] res = {ii/1000,ii%1000/100,ii%100/10,ii%10};
for(int j=0;j<base.length;j++) {
int a = base[j];
int b = res[res.length-1-j];
if(a!=b) {
flag = false;
break;
}
}
if(flag) {
System.out.println(i);
}
flag = true;
}
}
}