XJTU Summer Holiday Test 1(Divisibility by Eight-8的倍数)
来源:互联网 发布:js点击按钮执行ctrl c 编辑:程序博客网 时间:2024/06/05 00:35
8*125=1000 故一个数是8的倍数当且仅当它末尾3个数是8的倍数
所以答案最多有3位,暴力即可
#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<functional>#include<iostream>#include<cmath>#include<cctype>#include<ctime>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p]) #define Lson (x<<1)#define Rson ((x<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define MAXN (100000+10)typedef long long ll;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}char s[MAXN];int main(){//freopen("I.in","r",stdin);//freopen(".out","w",stdout);scanf("%s",s);char *pch=strstr(s,"AB");if (pch!=NULL&&strstr(pch+2,"BA")!=NULL){cout<<"YES"<<endl;return 0;}pch=strstr(s,"BA");if (pch!=NULL&&strstr(pch+2,"AB")!=NULL){cout<<"YES"<<endl;return 0;}cout<<"NO"<<endl;return 0;}
0 0
- XJTU Summer Holiday Test 1(Divisibility by Eight-8的倍数)
- XJTU Summer Holiday Test 1(Brackets in Implications-构造)
- Divisibility by Eight
- Codeforces550C:Divisibility by Eight
- Divisibility by Eight
- C. Divisibility by Eight
- Divisibility by Eight
- Codeforces Divisibility by Eight
- CF C. Divisibility by Eight
- [CodeForces550C]Divisibility by Eight[数学]
- CODEFORCES 550 C. Divisibility by Eight
- codeforces 550C Divisibility by Eight(数学题)
- 550C Divisibility by Eight(还是枚举)
- CodeForces 550C Divisibility by Eight(枚举)
- Codeforces 550 C. Divisibility by Eight
- codeforce 550c Divisibility by Eight (DFS)
- 【codeforces 550C】Divisibility by Eight
- cf 550c Divisibility by Eight 【规律】
- 【线段树】 HDOJ 5283 Senior's Fish
- shell学习四十二天----使用touch更新文件时间
- HDU 2199 二分查找
- [leetcode] 147.Insertion Sort List
- 交换排序之冒泡排序
- XJTU Summer Holiday Test 1(Divisibility by Eight-8的倍数)
- 最短路径—Dijkstra算法和Floyd算法
- Spark
- 快速使用svn仓库
- cocos2dx三角函数与小球
- iOS基础(foundation)-常用结构体
- 软件测试,想说爱你不容易
- scala的for循环 :枚举的“瑞士军刀”
- ORACLE_SQL语句总结