题解列表

筛选

题解 1998: 提成计算

解题思路:注意事项:参考代码:#include#includeintmain(){intn;doublesum=0;scanf("%d",&n);if(n<=100000){sum=n*0.1;}elseif(n<=400000){sum=10000+(n-1000

题解 2544: N以内累加求和

解题思路:注意事项:参考代码:#include#include#includeintmain(){intn,sum=0,i;scanf("%d",&n);for(i=0;i<=n;i++){sum+=i;}printf("%d",

题解 1859: 与2无关的数

解题思路:注意事项:参考代码:#include#include#includeintmain(){intn,sum=0,i;scanf("%d",&n);for(i=0;i<=n;i++){if(i%2!=0&&i%10!=2&&i/10%10!=