https://codeforces.com/contest/1208
f[0]=a, f[1]=b, f[n]=f[n-1]^f[n-2]， 求f[n]
可以发现，3个一周期
1 #include<iostream>
2 #include<sstream>
3 #include<fstream>
4 #include<algorithm>
5 #include<cstring>
6 #include<iomanip>
7 #include<cstdlib>
8 #include<cctype>
9 #include<vector>
10 #include<string>
11 #include<cmath>
12 #include<ctime>
13 #include<stack>
14 #include<queue>
15 #include<map>
16 #include<set>
17 #define mem(a,b) memset(a,b,sizeof(a))
18 #define random(a,b) (rand()%(b-a+1)+a)
19 #define ll long long
20 #define ull unsigned long long
21 #define e 2.71828182
22 #define Pi acos(-1.0)
23 #define ls(rt) (rt<<1)
24 #define rs(rt) (rt<<1|1)
25 #define lowbit(x) (x&(-x))
26 using namespace std;
27 int read()
28 {
29 int s=1,x=0;
30 char ch=getchar();
31 while(!isdigit(ch)) {if(ch=='-') s=-1;ch=getchar();}
32 while(isdigit(ch)) {x=10*x+ch-'0';ch=getchar();}
33 return x*s;
34 }
35
36 int main()
37 {
38 int test=read(),c;
39 while(test--)
40 {
41 int a=read(),b=read(),n=read();
42 c=a^b;
43 if(n%3==0) printf("%d\n",a);
44 else if(n%3==1) printf("%d\n",b);
45 else printf("%d\n",c);
46 }
47 }
给定n个数，求最短区间[l,r]使得去掉[l,r]区间内的数使得剩下的数两两互不相同。
由于数据范围不大(n<=2000)，时间又给了两秒，自然想到暴力
1 #include<iostream>
2 #include<sstream>
3 #include<fstream>
4 #include<algorithm>
5 #include<cstring>
6 #include<iomanip>
7 #include<cstdlib>
8 #include<cctype>
9 #include<vector>
10 #include<string>
11 #include<cmath>
12 #include<ctime>
13 #include<stack>
14 #include<queue>
15 #include<map>
16 #include<set>
17 #define mem(a,b) memset(a,b,sizeof(a))
18 #define random(a,b) (rand()%(b-a+1)+a)
19 #define ll long long
20 #define ull unsigned long long
21 #define e 2.71828182
22 #define Pi acos(-1.0)
23 #define ls(rt) (rt<<1)
24 #define rs(rt) (rt<<1|1)
25 #define lowbit(x) (x&(-x))
26 using namespace std;
27 const int MAXN=2e3+5;
28 int n,a[MAXN];
29 map<int,int> tmp,pool;
30 int read()
31 {
32 int s=1,x=0;
33 char ch=getchar();
34 while(!isdigit(ch)) {if(ch=='-') s=-1;ch=getchar();}
35 while(isdigit(ch)) {x=10*x+ch-'0';ch=getchar();}
36 return x*s;
37 }
38 /*void showtmp()
39 {
40 cout<<"tmp:\n";
41 for(auto &it:tmp)
42 cout<<it.first<<':'<<it.second<<endl;
43 }
44 void showpool()
45 {
46 cout<<"pool:\n";
47 for(auto &it:pool)
48 cout<<it.first<<':'<<it.second<<endl;
49 }*/
50 int main()
51 {
52 int n=read();
53 for(int i=1;i<=n;++i)
54 {
55 a[i]=read();
56 if(tmp.count(a[i])) tmp[a[i]]++;
57 else tmp[a[i]]=1;
58 }
59
60 int uni=tmp.size();
61 if(uni==n) return 0*printf("%d",0);
62 bool flag=false;
63 int ans=-1,lt,lp;//lt保存枚举[l,r]区间的长度，lp是剩下的长度
64 //tmp保存枚举的数及相应的个数，pool保存剩下的数及相应的个数
65 for(int i=n-uni;!flag;++i)//r
66 {
67 tmp.clear(),pool.clear(),lp=n-i,lt=i;
68 for(int j=1;j<=i;++j)
69 {
70 if(tmp.count(a[j])) tmp[a[j]]++;
71 else tmp[a[j]]=1;
72 }
73 for(int j=i+1;j<=n;++j)
74 {
75 if(pool.count(a[j])) pool[a[j]]++;
76 else pool[a[j]]=1;
77 }
78 if(pool.size()==lp)
79 {
80 flag=true;
81 ans=i;
82 }
83 for(int j=i+1;j<=n;++j)
84 {
85 if(pool.count(a[j-lt])) pool[a[j-lt]]++;
86 else pool[a[j-lt]]=1;
87 if(tmp.count(a[j])) tmp[a[j]]++;
88 else tmp[a[j]]=1;
89 if(pool[a[j]]==1) pool.erase(a[j]);
90 else pool[a[j]]--;
91 if(tmp[a[j-lt]]==1) tmp.erase(a[j-lt]);
92 else tmp[a[j-lt]]--;
93 /*cout<<"i: "<<i<<" j: "<<j<<" lt: "<<lt<<endl;
94 showtmp();showpool();*/
95 if(pool.size()==lp)
96 {
97 flag=true;
98 ans=i;break;
99 }
100
101 }
102
103 }
104 cout<<ans<<endl;
105 }
还是学习下高效一点的写法吧：