A
1 /*Huyyt*/
2 #include<bits/stdc++.h>
3 using namespace std;
4 typedef long long ll;
5 const int mod = 998244353;
6 const int maxn = 1e3 + 5;
7 int main()
8 {
9 int T;
10 cin >> T;
11 while(T--)
12 {
13 ll x;
14 scanf("%lld",&x);
15 printf("%lld\n",x-1);
16 }
17 return 0;
18 }B
#include<bits/stdc++.h>
#define lc(x) (2*x)
#define rc(x) (2*x+1)
#define fi first
#define se second
using namespace std;
typedef long long ll;
const ll mod = 998244353;
const int maxn = 1e5+5;
const int maxm = 1e2+5;
int n,m,k;
vector<int> v[maxm];
int lit[maxn];
int near[maxm];
int main(){
int t;
scanf("%d",&t);
for(int quq=1;quq<=t;quq++){
scanf("%d%d%d",&n,&m,&k);
int i,j,a,b;
for(i=1;i<=m;i++)v[i].clear();
for(i=1;i<=k;i++){
scanf("%d%d",&a,&b);
v[b].push_back(a);
}
ll ans=0;
memset(lit,0,sizeof(lit));
for(i=1;i<=m;i++){
memset(near,0,sizeof(near));
for(j=1;j<=n;j++){
lit[j]++;
}
for(j=0;j<v[i].size();j++)lit[v[i][j]]=0;
for(j=1;j<=n;j++){
b = near[0];
for(a=1;a<=lit[j];a++){
ans+=(j-b);
b = max(b,near[a]);
}
near[lit[j]]=j;
}
}
printf("Case #%d: %lld\n",quq,ans);
}
}C
1 /*Huyyt*/
2 #include<bits/stdc++.h>
3 #define mem(a,b) memset(a,b,sizeof(a))
4 using namespace std;
5 typedef long long ll;
6 const int mod = 998244353;
7 const int maxn = 2e4 + 5;
8 int n, m;
9 int num[maxn];
10 int ren[205][20];
11 int sum[205];
12 int ans[205];
13 int cur, biggestnum;
14 int biggestman;
15 int numcur;
16 bool flag = true;
17 void get_ans()
18 {
19 for (int i = 1; i <= n; i++)
20 {
21 for (int j = 1; j <= 13; j++)
22 {
23 ans[i] += ren[i][j] * j;
24 }
25 }
26 return ;
27 }
28 int get_min(int x)
29 {
30 for (int i = 3; i <= 13; i++)
31 {
32 if (ren[x][i])
33 {
34 return i;
35 }
36 }
37 if (ren[x][1])
38 {
39 return 1;
40 }
41 if (ren[x][2])
42 {
43 return 2;
44 }
45 }
46 int get_next(int x)
47 {
48 if (x == 2)
49 {
50 return -1;
51 }
52 if (x >= 3 && x <= 12)
53 {
54 return (x + 1);
55 }
56 if (x == 1)
57 {
58 return 2;
59 }
60 if (x == 13)
61 {
62 return 1;
63 }
64 }
65 int change_man(int x)
66 {
67 if (x > n)
68 {
69 return 1;
70 }
71 return x;
72 }
73 void chouka(int x)
74 {
75 for (int i = x; i <= n && cur <= m; i++)
76 {
77 ren[i][num[cur++]]++;
78 sum[i]++;
79 }
80 for (int i = 1; i < x && cur <= m; i++)
81 {
82 ren[i][num[cur++]]++;
83 sum[i]++;
84 }
85 }
86 void go(int mannow)
87 {
88 if (!flag)
89 {
90 return ;
91 }
92 mannow = change_man(mannow);
93 if (mannow == biggestman)
94 {
95 chouka(biggestman);
96 return ;
97 }
98 if (biggestnum == 2)
99 {
100 chouka(biggestman);
101 return ;
102 }
103 int aim = get_next(biggestnum);
104 if (aim == -1)
105 {
106 chouka(biggestman);
107 return ;
108 }
109 if (ren[mannow][aim])
110 {
111 ren[mannow][aim]--;
112 sum[mannow]--;
113 if (sum[mannow] == 0)
114 {
115 flag = false;
116 get_ans();
117 return ;
118 }
119 biggestman = mannow, biggestnum = aim;
120 go(mannow + 1);
121 }
122 else
123 {
124 if (ren[mannow][2])
125 {
126 ren[mannow][2]--;
127 sum[mannow]--;
128 if (sum[mannow] == 0)
129 {
130 flag = false;
131 get_ans();
132 return ;
133 }
134 biggestman = mannow, biggestnum = 2;
135 go(mannow + 1);
136 }
137 else
138 {
139 go(mannow + 1);
140 }
141 }
142 }
143 int main()
144 {
145 int T;
146 cin >> T;
147 for (int cas = 1; cas <= T; cas++)
148 {
149 flag = true;
150 numcur = 1, cur = 1, biggestnum = -1;
151 mem(ren, 0), mem(ans, 0), mem(sum, 0);
152 scanf("%d %d", &n, &m);
153 for (int i = 1; i <= m; i++)
154 {
155 scanf("%d", &num[i]);
156 }
157 for (int i = 1; i <= n; i++)
158 {
159 int add = 0;
160 for (; cur <= m && add <= 4;)
161 {
162 //cout<<i<<" "<<num[cur]<<endl;
163 ren[i][num[cur++]]++;
164 sum[i]++;
165 add++;
166 }
167 }
168 biggestman = 1;
169 while (flag)
170 {
171 if (sum[biggestman] == 0)
172 {
173 get_ans();
174 flag = false;
175 continue;
176 }
177 biggestnum = get_min(biggestman);
178 ren[biggestman][biggestnum]--;
179 sum[biggestman]--;
180 if (sum[biggestman] == 0)
181 {
182 get_ans();
183 flag = false;
184 continue;
185 }
186 go(biggestman + 1);
187 }
188 printf("Case #%d:\n", cas);
189 for (int i = 1; i <= n; i++)
190 {
191 if (ans[i] == 0)
192 {
193 printf("Winner\n");
194 }
195 else
196 {
197 printf("%d\n", ans[i]);
198 }
199 }
200 }
201 return 0;
202 }D
1 /*
2 Author: LargeDumpling
3 Email: LargeDumpling@qq.com
4 Edit History:
5 2018-09-01 File created.
6 */
7
8 #include<iostream>
9 #include<cstdio>
10 #include<cstdlib>
11 #include<cstring>
12 #include<cmath>
13 #include<algorithm>
14 using namespace std;
15 const int MAXN=1050;
16 const double eps=1e-10;
17 int dcmp(const double &x) { if(fabs(x)<eps) return 0; return x<0?-1:1; }
18 typedef struct Poi Vec;
19 struct Poi
20 {
21 double x,y;
22 Poi(const double &a=0,const double &b=0):x(a),y(b) { }
23 Poi operator+(const Poi &P)const { return Poi(x+P.x,y+P.y); }
24 Poi operator-(const Poi &P)const { return Poi(x-P.x,y-P.y); }
25 Poi operator*(const double &P)const { return Poi(x*P,y*P); }
26 Poi operator/(const double &P)const { return Poi(x/P,y/P); }
27 bool operator<(const Poi &P)const { return !dcmp(x-P.x)?dcmp(y-P.y)<=0:dcmp(x-P.x)<=0; }
28 }PP[MAXN],NP[MAXN<<1],IP[MAXN];
29 struct Line
30 {
31 Poi Ps; Vec Dir;
32 double ang;
33 Line(const Poi &ps=Poi(),const Vec &dir=Vec()):Ps(ps),Dir(dir) { ang=atan2(Dir.y,Dir.x); }
34 bool operator<(const Line &L1)const { return ang<L1.ang; }
35 }lines[MAXN],q[MAXN];
36 double dOt(const Vec &V1,const Vec &V2) { return V1.x*V2.x+V1.y*V2.y; }
37 double cRoss(const Vec &V1,const Vec &V2) { return V1.x*V2.y-V1.y*V2.x; }
38 double lEnth(const Vec V) { return sqrt(dOt(V,V)); }
39 Vec Normal(const Vec &V) { return Vec(-V.y,V.x); }
40 Poi iNtersect(const Poi &P1,const Vec &V1,const Poi &P2,const Vec &V2)
41 { return P1+V1*(cRoss(V2,P1-P2)/cRoss(V1,V2)); }
42 Poi iNtersect(const Line &L1,const Line &L2)
43 { return iNtersect(L1.Ps,L1.Dir,L2.Ps,L2.Dir); }
44 int HPI(Line *L,int N,Poi *Pol)
45 {
46 int l,r,m=0;
47 sort(L,L+N);//按极角排序
48 q[l=r=0]=L[0];
49 for(int i=1;i<N;i++)
50 {
51 while(l<r&&dcmp(cRoss(L[i].Dir,IP[r-1]-L[i].Ps))<=0) r--; //新加入的直线可能是尾部的一些交点失效
52 while(l<r&&dcmp(cRoss(L[i].Dir,IP[l]-L[i].Ps))<=0) l++; //首部
53 q[++r]=L[i]; //加入
54 if(!dcmp(cRoss(q[r].Dir,q[r-1].Dir)))
55 { //对于平行直线要取靠左的
56 r--;
57 if(dcmp(cRoss(q[r+1].Dir,q[r].Ps-q[r+1].Ps))<0) q[r]=q[r+1];
58 }
59 if(l<r) IP[r-1]=iNtersect(q[r-1],q[r]); //如果队列中有至少两条线，则取交点
60 }
61 while(l<r&&dcmp(cRoss(q[l].Dir,IP[r-1]-q[l].Ps))<=0) r--; //后面一些交点可能实际上是无用的
62 if(r-l<2) return 0; //如果只有不到两条线，则失败了
63 IP[r]=iNtersect(q[l],q[r]); //将最后一条线和第一条线交起来
64 for(int i=l;i<=r;i++) Pol[m++]=IP[i];
65 Pol[m]=Pol[0];
66 return m;
67 }
68 double area(const Poi &P1,const Poi &P2,const Poi &P3)
69 {
70 return fabs(cRoss(P3-P1,P2-P1));
71 }
72 double calc(int n,Poi *P)
73 {
74 int l,r,mid1,mid2,tem;
75 double ans=0;
76 for(int i=0;i<=n;i++)
77 P[i+n]=P[i];
78 for(int i=0;i<n;i++)
79 for(int j=i+1;j<n;j++)
80 {
81 l=i; r=j;
82 while(l<r-1)
83 {
84 tem=(r-l+1)/3;
85 mid1=l+tem;
86 mid2=r-tem;
87 if(area(P[i],P[j],P[mid1])<area(P[i],P[j],P[mid2])) l=mid1;
88 else r=mid2;
89 }
90 ans=max(ans,max(area(P[i],P[j],P[l]),area(P[i],P[j],P[r])));
91 l=j; r=i+n;
92 while(l<r-1)
93 {
94 tem=(r-l+1)/3;
95 mid1=l+tem;
96 mid2=r-tem;
97 if(area(P[i],P[j],P[mid1])<area(P[i],P[j],P[mid2])) l=mid1;
98 else r=mid2;
99 }
100 ans=max(ans,max(area(P[j],P[i+n],P[l]),area(P[j],P[i+n],P[r])));
101 }
102 return ans;
103 }
104 int main()
105 {
106 int T_T,n,m;
107 double r;
108 Vec Nv;
109 scanf("%d",&T_T);
110 while(T_T--)
111 {
112 scanf("%d%lf",&n,&r);
113 for(int i=0;i<n;i++) scanf("%lf%lf",&PP[n-i-1].x,&PP[n-i-1].y);
114 PP[n]=PP[0];
115 for(int i=0;i<n;i++)
116 {
117 Nv=Normal(PP[i+1]-PP[i]);
118 Nv=Nv/lEnth(Nv);
119 lines[i]=Line(PP[i]+Nv*r,PP[i+1]-PP[i]);
120 }
121 m=HPI(lines,n,NP);
122 printf("%lf\n",calc(m,NP));
123 }
124 fclose(stdin);
125 fclose(stdout);
126 return 0;
127 }E
1 /*
2 Author: LargeDumpling
3 Email: LargeDumpling@qq.com
4 Edit History:
5 2018-09-01 File created.
6 */
7
8 #include<iostream>
9 #include<cstdio>
10 #include<cstdlib>
11 #include<cstring>
12 #include<cmath>
13 #include<queue>
14 #include<algorithm>
15 using namespace std;
16 const int MAXN=25;
17 struct jz
18 {
19 int u,S,cnt;
20 jz(const int &U=0,const int &SS=0,const int &C=0):u(U),S(SS),cnt(C) { }
21 };
22 bool aa;
23 int n,pre[MAXN];
24 long long a[MAXN],b[MAXN],f[1100000],ans=0;
25 bool exist[1100000],vis[1100000];
26 bool bb;
27 void SPFA()
28 {
29 int S=0;
30 long long cnt=0;
31 queue<jz> q;
32 f[0]=0;
33 for(int i=0;i<n;i++) if((pre[i]&S)==pre[i]
34 &&(!vis[S|(1<<i)]||f[S|(1<<i)]<f[S]+a[i]*(cnt+1LL)+b[i]))
35 {
36 f[S|(1<<i)]=f[S]+a[i]*(cnt+1LL)+b[i];
37 vis[S|(1<<i)]=true;
38 q.push(jz(i,S|(1<<i),cnt+1));
39 exist[S|(1<<i)]=true;
40 }
41 while(q.size())
42 {
43 S=q.front().S;
44 cnt=q.front().cnt;
45 q.pop();
46 exist[S]=false;
47 for(int i=0;i<n;i++) if((!((S>>i)&1))&&((pre[i]&S)==pre[i])
48 &&(!vis[S|(1<<i)]||f[S|(1<<i)]<f[S]+a[i]*(cnt+1LL)+b[i]))
49 {
50 f[S|(1<<i)]=f[S]+a[i]*(cnt+1LL)+b[i];
51 vis[S|(1<<i)]=true;
52 ans=max(ans,f[S|(1<<i)]);
53 if(!exist[S|(1<<i)])
54 {
55 exist[S|(1<<i)]=true;
56 q.push(jz(i,S|(1<<i),cnt+1));
57 }
58 }
59 }
60 return;
61 }
62 int main()
63 {
64 int pn,x;
65 scanf("%d",&n);
66 for(int i=0;i<n;i++)
67 {
68 scanf("%lld%lld%d",&a[i],&b[i],&pn);
69 while(pn--)
70 {
71 scanf("%d",&x); x--;
72 pre[i]|=(1<<x);
73 }
74 }
75 SPFA();
76 printf("%lld\n",ans);
77 fclose(stdin);
78 fclose(stdout);
79 return 0;
80 }F
G
1 /*
2 Author: LargeDumpling
3 Email: LargeDumpling@qq.com
4 Edit History:
5 2018-09-01 File created.
6 */
7
8 #include<iostream>
9 #include<cstdio>
10 #include<cstdlib>
11 #include<cstring>
12 #include<cmath>
13 #include<algorithm>
14 using namespace std;
15 const int MAXN=100050;
16 const int M=131072;
17 const int INF=2147483647;
18 int n,m;
19 int mbl[MAXN],d[M<<1];
20 int ans[MAXN][2];
21 void read1n(int &x)
22 {
23 char ch;
24 for(ch=getchar();ch<'0'||'9'<ch;ch=getchar())
25 if(ch==-1) return;
26 for(x=0;'0'<=ch&&ch<='9';ch=getchar())
27 x=(x<<1)+(x<<3)+ch-'0';
28 return;
29 }
30 void build()
31 {
32 for(int i=1;i<=n;i++)
33 d[i+M]=mbl[i];
34 for(int i=M-1;i;i--)
35 d[i]=min(d[i<<1],d[i<<1|1]);
36 return;
37 }
38 void del(int x)
39 {
40 mbl[x]=d[x+M]=INF;
41 for(int i=(x+M)>>1;i;i>>=1)
42 d[i]=min(d[i<<1],d[i<<1|1]);
43 return;
44 }
45 int query(int L,int R)
46 {
47 int ans=INF;
48 for(L=L+M-1,R=R+M+1;L<R-1;L>>=1,R>>=1)
49 {
50 if(!(L&1)) ans=min(ans,d[L^1]);
51 if(R&1) ans=min(ans,d[R^1]);
52 }
53 return ans;
54 }
55 void calc()
56 {
57 int nex,cnt,now,l,mid,r,left=0;
58 ans[0][0]=ans[0][1]=0;
59 build();
60 cnt=0;
61 for(int i=1;i<MAXN;i++)
62 {
63 if(cnt==n)
64 {
65 ans[i][0]=ans[i-1][0];
66 ans[i][1]=ans[i-1][1];
67 continue;
68 }
69 left+=m;
70 now=1;
71 while(true)
72 {
73 if(mbl[now]<=left) nex=now;
74 else if(left<query(now,n)) nex=n+1;
75 else
76 {
77 l=now; r=n;
78 while(l<r-1)
79 {
80 mid=(l+r)>>1;
81 if(query(now,mid)<=left) r=mid;
82 else l=mid;
83 }
84 nex=r;
85 }
86 if(nex==n+1) break;
87 left-=mbl[nex];
88 cnt++;
89 del(nex);
90 now=nex;
91 }
92 ans[i][0]=cnt;
93 ans[i][1]=left;
94 }
95 }
96 int main()
97 {
98 int Q,x;
99 read1n(n); read1n(m);
100 for(int i=1;i<=n;i++)
101 scanf("%d",&mbl[i]);
102 calc();
103 read1n(Q);
104 while(Q--)
105 {
106 read1n(x);
107 printf("%d %d\n",ans[x][0],ans[x][1]);
108 }
109 fclose(stdin);
110 fclose(stdout);
111 return 0;
112 }H
I
1 /*
2 Author: LargeDumpling
3 Email: LargeDumpling@qq.com
4 Edit History:
5 2018-09-01 File created.
6 */
7
8 #include<iostream>
9 #include<cstdio>
10 #include<cstdlib>
11 #include<cstring>
12 #include<cmath>
13 #include<algorithm>
14 using namespace std;
15 const int MAXN=2000050;
16 const int MAXC=10;
17 const long long mod=1000000007LL;
18 char str[MAXN];
19 int ch[MAXN][MAXC],len[MAXN],fail[MAXN],last,sz;
20 long long num[MAXN],ans,p[MAXN];
21 int getfail(char T[],int x,int i)
22 {
23 while(T[i-len[x]-1]!=T[i]) x=fail[x];
24 return x;
25 }
26 void init()
27 {
28 memset(ch[0],last=0,sizeof(ch[0]));//last为当前的最长后缀回文
29 memset(ch[1],0,sizeof(ch[1]));
30 len[0]=0; len[1]=-1;
31 fail[0]=1;
32 num[0]=num[1]=0;
33 sz=1;
34 ans=0;
35 return;
36 }
37 void insert(char T[])
38 {
39 int lenth=strlen(T),cur=0;
40 for(int i=0;i<lenth;i++)
41 {
42 cur=getfail(T,last,i);
43 if(!ch[cur][T[i]-'0'])
44 {
45 fail[++sz]=ch[ getfail(T,fail[cur],i) ][T[i]-'0'];
46 ch[cur][T[i]-'0']=sz;
47 memset(ch[sz],0,sizeof(ch[sz]));
48 len[sz]=len[cur]+2;
49 if(cur==1) num[sz]=T[i]-'0';
50 else num[sz]=(num[cur]*10LL%mod+p[len[cur]+1]*(T[i]-'0')%mod+(T[i]-'0'))%mod;
51 ans=(ans+num[sz])%mod;
52 }
53 last=ch[cur][T[i]-'0'];
54 }
55 return;
56 }
57 void pre_calc()
58 {
59 p[0]=1;
60 p[1]=10LL;
61 for(int i=2;i<MAXN;i++) p[i]=p[i-1]*p[1]%mod;
62 return;
63 }
64 int main()
65 {
66 init();
67 pre_calc();
68 scanf("%s",str);
69 insert(str);
70 printf("%lld\n",ans);
71 fclose(stdin);
72 fclose(stdout);
73 return 0;
74 }J
1 #include<bits/stdc++.h>
2 #define lc(x) (2*x)
3 #define rc(x) (2*x+1)
4 #define fi first
5 #define se second
6 using namespace std;
7 typedef long long ll;
8
9 const ll mod = 998244353;
10 const int maxn = 2e7+5;
11 const int maxm = 1e4;
12 int num[maxn],is[maxn];
13 int n;
14
15 void yuchuli(){
16 int i,j,a,b;
17 for(i=2;;i++){
18 if(i*i>=maxn)break;
19 a=i*i;
20 for(j=a;j<maxn;j+=a)is[j]=1;
21 }
22 for(i=1;i<maxn;i++)num[i]=num[i-1]+1-is[i];
23 }
24
25 int main(){
26 int t;
27 yuchuli();
28 scanf("%d",&t);
29 while(t--){
30 scanf("%d",&n);
31 int i,j,a,b;
32 ll ans=0;
33 for(i=1;i<=maxm;i++){
34 if(is[i])continue;
35 j=n/i;
36 ans += num[j];
37 if(j>maxm)ans+=num[j]-num[maxm];
38 }
39 printf("%lld\n",ans);
40 }
41 }K
L
1 /*
2 Author: LargeDumpling
3 Email: LargeDumpling@qq.com
4 Edit History:
5 2018-09-01 File created.
6 */
7
8 #include<iostream>
9 #include<cstdio>
10 #include<cstdlib>
11 #include<cstring>
12 #include<cmath>
13 #include<queue>
14 #include<algorithm>
15 using namespace std;
16 const int MAXN=100050;
17 const int MAXM=200050;
18 struct jz
19 {
20 int u;
21 long long dis;
22 jz(const int &U=0,const long long &D=0):u(U),dis(D) { }
23 bool operator<(const jz &X)const { return dis==X.dis?u>X.u:dis>X.dis; }
24 };
25 bool aaa;
26 int T_T,n,m,k;
27 int fir[MAXN*15],eNd[MAXM*25],nxt[MAXM*25],ed=0;
28 bool vis[MAXN*15];
29 long long len[MAXM*25],dis[MAXN*15],ans;
30 bool bbb;
31 void addedge(int u,int v,int w)
32 {
33 eNd[++ed]=v;
34 nxt[ed]=fir[u];
35 len[ed]=w;
36 fir[u]=ed;
37 return;
38 }
39 void read1n(int &x)
40 {
41 char ch=getchar();
42 while(ch<'0'||'9'<ch) ch=getchar();
43 for(x=0;'0'<=ch&&ch<='9';ch=getchar())
44 x=(x<<1)+(x<<3)+ch-'0';
45 return;
46 }
47 void init()
48 {
49 memset(fir,ed=0,sizeof(fir));
50 memset(dis,-1,sizeof(dis));
51 memset(vis,false,sizeof(vis));
52 return;
53 }
54 void Hijkstra()
55 {
56 int u;
57 priority_queue<jz> q;
58 dis[1]=0;
59 q.push(jz(1,0));
60 while(q.size())
61 {
62 while(q.size()&&(vis[q.top().u]||dis[q.top().u]!=q.top().dis))
63 q.pop();
64 if(!q.size()) break;
65 u=q.top().u; q.pop();
66 vis[u]=true;
67 for(int i=fir[u];i;i=nxt[i]) if(dis[eNd[i]]==-1||dis[u]+len[i]<dis[eNd[i]])
68 {
69 dis[eNd[i]]=dis[u]+len[i];
70 q.push(jz(eNd[i],dis[eNd[i]]));
71 }
72 }
73 return;
74 }
75 int Rand()
76 {
77 return (rand()<<15)+rand();
78 }
79 int main()
80 {
81 int u,v,w;
82 read1n(T_T);
83 //T_T=5;
84 while(T_T--)
85 {
86 init();
87 read1n(n);
88 //n=100000;
89 read1n(m);
90 //m=200000;
91 read1n(k);
92 //k=10;
93 for(int i=1;i<=m;i++)
94 {
95 read1n(u);
96 //u=Rand()%n+1;
97 read1n(v);
98 //v=Rand()%n+1;
99 read1n(w);
100 //w=1000000000;
101 for(int j=0;j<=k;j++)
102 {
103 addedge(u+j*n,v+j*n,w);
104 if(j<k) addedge(u+j*n,v+(j+1)*n,0);
105 }
106 }
107 Hijkstra();
108 ans=dis[n];
109 for(int i=1;i<=k;i++)
110 ans=min(ans,dis[(i+1)*n]);
111 printf("%lld\n",ans);
112 }
113 fclose(stdin);
114 fclose(stdout);
115 return 0;
116 }
来源：oschina
链接：https://my.oschina.net/u/4263629/blog/3840227