Codeforces Round #262 (Div. 2)
来源:互联网 发布:mac自带截图快捷键 编辑:程序博客网 时间:2024/06/06 18:28
Little Dima misbehaved during a math lesson a lot and the nasty teacher Mr. Pickles gave him the following problem as a punishment.
Find all integer solutions x (0 < x < 109) of the equation:
where a, b,c are some predetermined constant values and functions(x) determines the sum of all digits in the decimal representation of numberx.
The teacher gives this problem to Dima for each lesson. He changes only the parameters of the equation:a, b,c. Dima got sick of getting bad marks and he asks you to help him solve this challenging problem.
The first line contains three space-separated integers: a, b, c (1 ≤ a ≤ 5; 1 ≤ b ≤ 10000; - 10000 ≤ c ≤ 10000).
Print integer n — the number of the solutions that you've found. Next printn integers in the increasing order — the solutions of the given equation. Print only integer solutions that are larger than zero and strictly less than109.
3 2 8
310 2008 13726
1 2 -18
0
2 2 -1
41 31 337 967 题目意思是:给出a,b,c,求满足等式 x=b*s(x)^a+c 的所有x(s(x)表示x的十进制各数位之和)。没有就输出0。一开始一点思路都没有,后来仔细想了想,s(x)最大只有81,所以可以从s(x)开始枚举就可以了。#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;int ss[1000];int main(){ int a,b,c; int k=0; scanf("%d%d%d",&a,&b,&c); for(int i=1;i<=81;i++) { int bbb=b; for(int j=1;j<=a;j++) bbb*=i; int ans=c+bbb; //cout<<ans<<" "; int tans=ans; //cout<<tans<<" "; if(ans<=0) continue; if(ans>=1000000000) break; int sum=0; while(ans){ sum+=ans%10; ans/=10; } if(sum==i){// cout<<sum<<endl; ss[k++]=tans;//printf("%d ",tans); } } if(ss[0]==0) printf("0\n"); else{ printf("%d\n",k); for(int i=0;i<k;i++) { if(i==k-1) printf("%d\n",ss[i]); else printf("%d ",ss[i]); } } return 0;}
- Codeforces Round #262 (Div. 2)
- Codeforces Round #262 (Div. 2)
- Codeforces Round #262 (Div. 2)
- Codeforces Round #262 (Div. 2)
- Codeforces Round #262 (Div. 2)
- Codeforces Round #262 (Div. 2)
- Codeforces Round #262 (Div. 2)总结
- Codeforces Round #262 (Div. 2) A
- Codeforces Round #262 (Div. 2) B
- Codeforces Round #262 (Div. 2) C
- Codeforces Round #262 (Div. 2) 总结:二分
- Codeforces Round #262 (Div. 2)解题报告
- Codeforces Round #262 (Div. 2) C
- Codeforces Round #262 (Div. 2)思维训练
- Codeforces Round #262 (Div. 2)总结
- Codeforces Round #262 (Div. 2) 题解
- Codeforces Round #262 (Div. 2) B
- Codeforces Round #262 (Div. 2) C
- 在Linux下编写C程序,怎么检查程序是否有内存泄漏?
- django+celery+RabbitMQ 环境配置
- Java中Properties类的使用
- 客户端封装数据为json格式的数据提交到服务端的方法
- PHP的$_SERVER['HTTP_HOST']获取服务器地址功能详解,$_SERVER['HTTP_X_FORWARDED_HOST']
- Codeforces Round #262 (Div. 2)
- 阿迪达斯智能足球使用Nordic Semiconductor蓝牙智能无线技术
- 【COCOS2DX-BOX2D游戏开发之三】 读取tiledmap的tmx阻挡
- 基于事件的 NIO 多线程服务器
- ORA-00376: 此时无法读取文件
- Oracle cloud control 12c 如何修改sysman密码
- 二十六个月Android学习工作总结
- Android之旅十六 android中各种资源的使用
- 【CocoStudio游戏开发之一】制作多分辨率UI布局