不要挡住我的阳光

#include"iostream"#include"algorithm"#include"vector"#include"map"#include"string"#include"deque"#include"queue"#include"stack"#include"cstdio"#include"ctype.h"#include"cmath"#include"set"

#include"map"#include"iostream"#include"vector"#include"stdio.h"using namespace std;typedef long long ll;const int prime=14997;const int maxn=100000+10;vectorHash[prime];int g[maxn][6];int c

#include"iostream"#include"string"#include"cstring"#include"algorithm"#include"cstdio"using namespace std;struct node{ node *lchild,*rchild; char val;};string f(string a,int b,int e)

精度很重要,看到大数字就要注意是否要使用long long  或者是 __int64#include"iostream"#include"cstdio"#include"algorithm"#include"vector"#include"string"#include"stack"using namespace std;const int maxn=100000+10;__i

#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include inline in

#include #include #include #define ls L,M,rt<<1#define rs M+1,R,rt<<1|1using namespace std;typedef long long ll;const int maxn = 111110;ll tree[maxn<<2];ll add[maxn<<2];void pushup(int rt)

这个map我也是醉了。这个 主要是 那个   不同gcd值得区间数预处理难弄。a[1...i]=a[1...i-1]+a[i...i];  当前区间:不同的gcd值相对应的区间数=之前区间:不同的gcd值相对应的区间数+gcd(num[i],之前区间的某个gcd);#include"iostream"#include"vector"#include"set"#include"qu

//DFS#include #include #include using namespace std;vector > V; //邻接表vector visit;int n;//点数int m;//边数vector ans;void dfs(int n){ ans.push_back(n); visit[n] = 1; for(int j=0;j<V[

#include"iostream"#include"vector"#include"stack"using namespace std;const int INF = 1111111;struct Edge{ //end，len（start--->end) int ed; int len;};struct Node{ int len; int st;

#include"iostream"#include"stack"#include"vector"using namespace std;vector > G;//采用邻接表保存图vector degree; //度数表vector is_in_stack;//判断是否入过栈vector ans; //保存答案int n,m; /

#include"iostream"using namespace std;const int MAXN = 5; //队列中最大能够存储的元素个数int Queue[MAXN]; //队列int Front,Rail; //队首，队尾int push(int e){ //插入队首 Queue[Rail] = e; Rail = (Rail+1)%MAXN;}

#include"iostream"#include"cstdio"#include"string"#include"algorithm"using namespace std;int main(){ //freopen("a.txt","r",stdin); char a[1000]; char b[1000]; while(cin>>a>>b) { int i,

#include"stdio.h"#include"iostream"#include"stack"using namespace std;int main(){ //freopen("a.txt","r",stdin); int n,pos,a[1010]; while(cin>>n,n) { while(cin>>a[1],a[1])

sumTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1372    Accepted Submission(s): 572Problem DescriptionGiven a sequence, you're

#include"iostream"#include"cstdio"using namespace std;const int maxn=100000+10; //数组开大点struct trie{ int val[maxn]; int ch[maxn][10]; int sz; trie() { sz=1; memset(ch[0],0,sizeof(ch[0]))

#include"iostream"#include"cstdlib"using namespace std;const int MAXN = 199;int arr[MAXN];int n;void swap(int i,int j){ int t = arr[i];arr[i] = arr[j];arr[j] = t;}void quick_sort(int left