每日一题:第3天

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;    
        }
    }
}

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转载自www.cnblogs.com/wangbaby/p/10668450.html