C题大水题，欧拉筛筛下素数，然后在线处理一下普通素数。

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 #define scan(i) scanf("%d",&i)
 4 #define scanl(i) scanf("%lld",&i)
 5 #define scand(i) scanf("%lf",&i)
 6 #define pf printf
 7 #define f(i,a,b) for(int i=a;i<=b;i++)
 8 const int N=1e7;
 9 int phi[N+10],prime[N+10],tot,ans;
10 bool mark[N+10];
11 void getphi()    {
12    int i,j;
13    phi[1]=1;
14    for(i=2;i<=N;i++){
15        if(!mark[i]){
16              prime[++tot]=i;//筛素数的时候首先会判断i是否是素数。
17              phi[i]=i-1;//当 i 是素数时 phi[i]=i-1
18              }
19        for(j=1;j<=tot;j++){
20           if(i*prime[j]>N)  break;
21           mark[i*prime[j]]=1;//确定i*prime[j]不是素数
22           if(i%prime[j]==0){//接着我们会看prime[j]是否是i的约数
23              phi[i*prime[j]]=phi[i]*prime[j];break;
24           }
25           else phi[i*prime[j]]=phi[i]*(prime[j]-1);//其实这里prime[j]-1就是phi[prime[j]]，利用了欧拉函数的积性
26        }
27    }
28 }
29 //调用时，给出getphi();即可，mark[i]是false说明是素数，否则是合数
30 
31 bool isPrime(int num)
32 {
33     if (num == 2 || num == 3)
34     {
35         return true;
36     }
37     if (num % 6 != 1 && num % 6 != 5)
38     {
39         return false;
40     }
41     for (int i = 5; i*i <= num; i += 6)
42     {
43         if (num % i == 0 || num % (i+2) == 0)
44         {
45             return false;
46         }
47     }
48     return true;
49 }
50 
51 using namespace std;
52 int n,m,T;
53 int l,r;
54 int main()
55 {
56     getphi();
57     scan(T);
58     f(kk,1,T){
59         scanf("%d%d",&l,&r);
60         if(r<=1000){
61             int cnt=0;
62             f(i,l,r){
63                 if(mark[i]==false) cnt++;
64             }
65             if(cnt*3<r-l+1) pf("Yes\n");
66             else pf("No\n");
67         }
68         else if(r-l+1>=12){
69             pf("Yes\n");
70         }
71         else{
72             int cnt=0;
73             f(i,l,r){
74                 if(isPrime(i)) cnt++;
75             }
76             if(cnt*3<r-l+1) pf("Yes\n");
77             else pf("No\n");
78         }
79     }
80     return 0;
81 }

F题小水题，离线打表O(1)查询。要预处理一下立方数，会快一点。

  1 #include <bits/stdc++.h>
  2 #define ll long long
  3 #define scan(i) scanf("%d",&i)
  4 #define scanl(i) scanf("%lld",&i)
  5 #define scand(i) scanf("%lf",&i)
  6 #define pf printf
  7 #define f(i,a,b) for(int i=a;i<=b;i++)
  8 const double eps=1e-8;
  9 using namespace std;
 10 int ans[604]={
 11 0,0,0,
 12 0,0,1,
 13 0,1,1,
 14 1,1,1,
 15 -5001,0,0,
 16 -5001,0,0,
 17 -1,-1,2,
 18 0,-1,2,
 19 0,0,2,
 20 0,1,2,
 21 1,1,2,
 22 297,-641,619,
 23 7,-11,10,
 24 -5001,0,0,
 25 -5001,0,0,
 26 2,-1,2,
 27 0,2,2,
 28 1,2,2,
 29 75,-218,215,
 30 0,-2,3,
 31 1,-2,3,
 32 28,-86,85,
 33 -5001,0,0,
 34 -5001,0,0,
 35 2,2,2,
 36 1839,-2683,2357,
 37 0,-1,3,
 38 0,0,3,
 39 0,1,3,
 40 1,1,3,
 41 -5001,0,0,
 42 -5001,0,0,
 43 -5001,0,0,
 44 -5001,0,0,
 45 2,-1,3,
 46 0,2,3,
 47 1,2,3,
 48 0,-3,4,
 49 1,-3,4,
 50 -5001,0,0,
 51 -5001,0,0,
 52 -5001,0,0,
 53 -5001,0,0,
 54 2,2,3,
 55 -5,-7,8,
 56 2,-3,4,
 57 3,-2,3,
 58 6,-8,7,
 59 -2,-2,4,
 60 -5001,0,0,
 61 -5001,0,0,
 62 602,-796,659,
 63 -5001,0,0,
 64 3,-1,3,
 65 0,3,3,
 66 1,3,3,
 67 0,-2,4,
 68 1,-2,4,
 69 -5001,0,0,
 70 -5001,0,0,
 71 -1,-4,5,
 72 0,-4,5,
 73 2,3,3,
 74 0,-1,4,
 75 0,0,4,
 76 0,1,4,
 77 1,1,4,
 78 -5001,0,0,
 79 -5001,0,0,
 80 2,-4,5,
 81 11,-21,20,
 82 2,-1,4,
 83 0,2,4,
 84 1,2,4,
 85 -5001,0,0,
 86 -5001,0,0,
 87 -5001,0,0,
 88 -5001,0,0,
 89 26,-55,53,
 90 -19,-33,35,
 91 2,2,4,
 92 3,3,3,
 93 847,-1317,1188,
 94 3,-2,4,
 95 -5001,0,0,
 96 -5001,0,0,
 97 -5001,0,0,
 98 -1972,-4126,4271,
 99 3,-4,5,
100 6,-7,6,
101 3,-1,4,
102 0,3,4,
103 1,3,4,
104 -5,-5,7,
105 -5001,0,0,
106 -5001,0,0,
107 14,-22,20,
108 17,-22,18,
109 0,-3,5,
110 2,3,4,
111 -3,-6,7,
112 4,-3,4,
113 118,-239,229,
114 -5001,0,0,
115 -5001,0,0,
116 -4,-7,8,
117 2,-3,5,
118 947,-1309,1117,
119 -948,-1165,1345,
120 146,-1019,1018,
121 -5001,0,0,
122 148,-1040,1039,
123 -5001,0,0,
124 -5001,0,0,
125 -5001,0,0,
126 8,-12,11,
127 -1,-2,5,
128 0,-2,5,
129 3,3,4,
130 -2,-6,7,
131 4,-2,4,
132 -5001,0,0,
133 -5001,0,0,
134 -1,-1,5,
135 0,-1,5,
136 0,0,5,
137 0,1,5,
138 1,1,5,
139 0,4,4,
140 1,4,4,
141 -5001,0,0,
142 -5001,0,0,
143 2,-1,5,
144 0,2,5,
145 1,2,5,
146 2,-6,7,
147 2,4,4,
148 -9,-11,13,
149 -77,-86,103,
150 -5001,0,0,
151 -5001,0,0,
152 2,2,5,
153 -3,-7,8,
154 -5001,0,0,
155 3,-2,5,
156 -7,-8,10,
157 108,-149,127,
158 1528,-2366,2131,
159 -5001,0,0,
160 -5001,0,0,
161 260,-367,317,
162 3,-1,5,
163 0,3,5,
164 1,3,5,
165 3,-6,7,
166 3,4,4,
167 -5001,0,0,
168 -5001,0,0,
169 -5001,0,0,
170 80,-130,119,
171 2,3,5,
172 20,-31,28,
173 4,-3,5,
174 226,-1134,1131,
175 -45,-47,58,
176 -5001,0,0,
177 -5001,0,0,
178 -5001,0,0,
179 56,-172,170,
180 0,-7,8,
181 1,-7,8,
182 67,-87,71,
183 -5001,0,0,
184 -5001,0,0,
185 7,-8,7,
186 -5001,0,0,
187 -5001,0,0,
188 2,-7,8,
189 -10,-13,15,
190 3,3,5,
191 -5001,0,0,
192 4,-2,5,
193 9,-14,13,
194 10,-17,16,
195 -5001,0,0,
196 -5001,0,0,
197 5,-4,5,
198 27,-58,56,
199 4,-1,5,
200 0,4,5,
201 1,4,5,
202 4,-6,7,
203 4,4,4,
204 -5001,0,0,
205 -5001,0,0,
206 -5001,0,0,
207 3,-7,8,
208 2,4,5,
209 148,-195,161,
210 15,-24,22,
211 73,-114,103};
212 int l,r;
213 struct p{
214     int a;int b;int c;
215     p(){}
216     p(int aa,int bb,int cc){
217         a=aa;
218         b=bb;
219         c=cc;
220     }
221 }vec[202];
222 int main()
223 {
224     /*freopen("in.txt","r",stdin);
225     f(i,0,200){
226         stringstream ss;
227         string s;
228         getline(cin,s);
229         if(s.length()<=2){
230             cout<<"-5001,0,0,"<<endl;
231             continue;
232         }
233         //scanf("%d",&vec[i].a);
234         ss.str(s);
235         ss>>vec[i].a;
236             //scanf("%d%d",&vec[i].b,&vec[i].c);
237             ss>>vec[i].b>>vec[i].c;
238             cout<<vec[i].a<<","<<vec[i].b<<","<<vec[i].c<<",\n";
239     }*/
240     int T,n;
241     scan(T);
242     f(kk,1,T){
243         scan(n);
244         if(ans[3*n]==-5001){puts("impossible");}
245         else pf("%d %d %d\n",ans[3*n],ans[3*n+1],ans[3*n+2]);
246     }
247 }

A题数论题。已将结论整理在ACM-数论：一些零碎的数论定理中。

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 #define scan(i) scanf("%d",&i)
 4 #define scanl(i) scanf("%lld",&i)
 5 #define scand(i) scanf("%lf",&i)
 6 #define pf printf
 7 #define f(i,a,b) for(int i=a;i<=b;i++)
 8 using namespace std;
 9 int T;
10 ll l,r,s;
11 ll cal(ll x){
12     switch(x%4){
13         case 0:return x;
14         case 1:return 1;
15         case 2:return x+1;
16         case 3:return 0;
17     }
18 }
19 int main()
20 {
21     scan(T);
22     f(kk,1,T){
23         scanf("%lld%lld%lld",&l,&r,&s);
24         ll lef=(l+1)/2*2;
25         ll m=(lef+3)%4ll;
26         ll rig=(r-m)/4*4+m;
27         //cout<<lef<<" "<<rig<<endl;
28         if(rig>lef){
29             ll ans=0;
30             ll ansv=rig-lef+1;
31             //ll zuo=lef-l;
32             //ll you=r-rig;
33             for(ll i=lef;i>=l;i--){
34                 ll ans1=ans;
35                 for(ll j=rig;j<=r;j++){
36                     //if(i==lef&&j==rig) continue;
37                     if(j!=rig) ans1^=j;
38                     if(ans1<=s){
39                         ansv=max(ansv,j-i+1);
40                     }
41                 }
42                 ans^=(i-1);
43             }
44             //cout<<ansv<<endl;
45             pf("%lld\n",ansv);
46         }
47         else{
48             ll ans=-1;
49             //cout<<s<<endl;
50             for(ll i=l;i<=r;i++){
51                 for(ll j=i;j<=r;j++){
52                     if((cal(i-1)^cal(j))<=s){
53                         ans=max(ans,j-i+1);
54                         //cout<<i-1<<" "<<j<<" "<<(cal(i-1)^cal(j))<<" "<<s<<endl;
55                     }
56                 }
57             }
58             //cout<<ans<<endl;
59             pf("%lld\n",ans);
60         }
61     }
62     return 0;
63 }

M题关于树。我一定要学会数据结构中的树，只靠那一点自欺欺人的数论和图论知识是没有办法独当一面的。只有自己才最靠得住。加油啊。