2018ICPC 焦作 训练赛
Posted hugh-locke
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了2018ICPC 焦作 训练赛相关的知识,希望对你有一定的参考价值。
赛后总结:
先把所有的可做题放上线程,debug在线程上有题的时候永远下机找
宁愿机下多线程debug不要机下多线程空想题
A. 0:05:54 solved by hl
温暖的签到
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read()int x = 0,f = 1;char c = getchar();while (c<‘0‘ || c>‘9‘)if (c == ‘-‘) f = -1;c = getchar(); while (c >= ‘0‘&&c <= ‘9‘)x = x * 10 + c - ‘0‘;c = getchar();return x*f; const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 110; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; int main() int T = read(); while(T--) int ans = 0; for(int i = 1; i <= 4 ; i ++) ans += (read() != 0); if(ans == 0) puts("Typically Otaku"); else if(ans == 1) puts("Eye-opener"); else if(ans == 2) puts("Young Traveller"); else if(ans == 3) puts("Excellent Traveller"); else puts("Contemporary Xu Xiake"); return 0;
B.3:57:18 solved by gbs
思维题
要打倒两个敌人,那么必然先打倒一个,后打倒另一个
设打倒第一个敌人 hp为hp_1 打倒其需要花费回合turn_1 第二个则是hp_2 turn_2 ,设turn_sum = turn_1 + turn_2
设F(x) = (1+x)*x/2; 可以理解为前x回合造成的总伤害(包括溢出伤害)
那么一定有F(turn_1)>=hp_1 F(turn_sum)>=hp_1 + hp_2 否则总伤害不够
有一条性质
计数列 1,2,3……n 的和为sum 那么对于任意的1<=k <=sum 那么一定存在一种取法,从数列中取出若干数,使和为k(一种取法:从n到1枚举,若可以取便取。直到和等于k)
类似的,在攻击第一个敌人时,可以取出若干次攻击来攻击第二个敌人,那么不会有伤害溢出。
由此容易推断出满足F(turn_1)>=hp_1 和F(turn_sum)>=hp_1 + hp_2 的最小 turn_1 和turn_sum即是最优解,可以用二分来查找
分先攻A 和先攻B两种情况讨论,就可以得到受到的伤害最小(存在turn_1 = turn_sum的情况,但这种情况不处理貌似对解题也没有影响)
麻烦的是要输出字典序最小的攻击方案,这里分先攻A和先攻B两种情况
首先要计算如果前turn_1轮全打第一个敌人的溢出伤害 over_damage1 打完两个敌人最多溢出的伤害为over_damage2
可以知道over_damage1 <turn_1 否则可以提前一轮打败第一个敌人
假设在前turn_1轮 期间对第二个敌人造成了 k1点伤害
那么k1>=over_damage1-over_damage2 否则打败后一个敌人的时间要延后一轮 不是最优解
那么有 over_damage1-over_damage2 <=k1 <=over_damage1
如果先攻A
若over_damage1-over_damage2<=0则 前turn_1轮全攻A 后面的轮全攻B
如果over_damage1-over_damage2>0,那么在前turn_1轮必须对B造成伤害 索性造成over_damage1的伤害,那么只要再第 over_damage1 轮攻击一次B即可,前面已经证明了over_damage1 <turn_1
此时B出现最靠后,字典序最小
如果先攻B
在前turn_1次攻击中 选择若干次攻击A
此时同样有over_damage1-over_damage2 <=k1 <=over_damage1
此时还要用到上的的函数F(x) = (1+x)*x/2
找到最大的t 使得F(t) >=over_damage1 ,如果不考虑over_damage1-over_damage2 <=k1这个条件
那么答案必然是若干个A+若干个B+若干个A的形式 例如如AAABBBBAAAAA (否则,如果出现A在两个B之间,则可以把这个A左移,使得字典序变小)
考虑到over_damage1-over_damage2 <=k1这个条件 在构造后如果不满足这个条件 ,那么把前面若干个连续的A最后一个A的位置右移即可
如果先攻A和先攻B,受到的总伤害相同,那么对于这两种情况都计算一下,然后比较字符串的大小
#include <iostream> #include<stack> #include<math.h> #include<stdlib.h> #include<string.h> #include<string> #include<ctime> #include<complex> #include<stdio.h> #include<algorithm> #include<map> #include<queue> using namespace std; typedef long long LL; const int mod =1e9+7; int hpa,hpb,atka,atkb; int hpab; int sum_turn; int a_turn; int a_re; int b_turn; int b_re; LL ap_damage; LL bp_damage; int sum_r; int give_up,give_down; int theover(int a) int left1 = 0; int right1 = 70000; LL ans; while(right1 -left1>1) int mid = (left1 +right1)/2; ans =1LL *mid*(mid+1)/2; if (ans >=a) right1 =mid; else left1 = mid; return right1; int theblow(int a) int left1 = 0; int right1 = 70000; int mid; LL ans; while(right1 -left1>1) int mid = (left1 +right1)/2; ans =1LL *mid*(mid+1)/2; if (ans <=a) left1 =mid; else right1 = mid; return left1; char ansa[100005]; char ansb[100005]; bool cmp1() for (int i=0; i<sum_turn; i++) if (ansa[i] < ansb[i]) return true; if (ansa[i] > ansb[i]) return false; return true; int main() int t; cin >>t; while(t--) scanf("%d%d%d%d",&hpa,&hpb,&atka,&atkb); hpab = hpa+hpb; sum_turn = theover(hpab); sum_r = 1LL * (sum_turn+1)*sum_turn/2 - hpab; a_turn = theover(hpa); a_re = 1LL *(a_turn +1)*a_turn /2-hpa; b_turn = theover(hpb); b_re = 1LL *(b_turn +1)*b_turn /2-hpb; ap_damage = 1LL*atka * a_turn + 1LL*sum_turn * atkb; bp_damage = 1LL*atkb * b_turn + 1LL*sum_turn * atka; bool if_a_first= false; bool if_b_c = false; if (a_turn == sum_turn) if_a_first = false; else if (b_turn == sum_turn) if_a_first = true; else if (ap_damage <= bp_damage) if_a_first = true; if (ap_damage == bp_damage) if_b_c = true; else if_a_first = false; LL fin_ans; if (if_a_first)//A first fin_ans = ap_damage; give_up = a_re; give_down = a_re - sum_r; for (int i=0; i<a_turn; i++) ansa[i] =‘A‘; for (int i=a_turn; i<sum_turn; i++) ansa[i] =‘B‘; ansa[sum_turn] = 0; if (give_down>0) ansa[give_up-1]=‘B‘; if (if_b_c || !if_a_first) fin_ans = bp_damage; give_up = b_re; give_down = b_re - sum_r; for (int i=0; i<b_turn; i++) ansb[i] =‘B‘; for (int i=b_turn; i<sum_turn; i++) ansb[i] =‘A‘; ansb[sum_turn] = 0; int dt1 = theblow(b_re); for (int i =0; i<dt1; i++) ansb[i] =‘A‘; int now_have = dt1 * (dt1+1)/2; if (now_have <give_down) now_have = give_down -now_have; ansb[dt1-1]=‘B‘; ansb[dt1-1+now_have]=‘A‘; cout<<fin_ans<<‘ ‘; if (if_b_c) if(cmp1()) printf("%s\n",ansa); else printf("%s\n",ansb); else if(if_a_first) printf("%s\n",ansa); else printf("%s\n",ansb); return 0; /* 111 5111111 11111116 12133215 251111111 111 5 15 5 25 5 14 5 25 5 13 5 25 5 12 5 25 3 14 14 5 5 14 14 5 5 */
C.3:01:00(-1) solved by zcz
用四个变量维护当前棋子的最左边,最右边,最上边,最下边
由八皇后棋子摆法的特性可知 只有同时存在于最左(右)和最上(下)的棋子才有可能重叠。
与此同时用两个变量表示棋盘的移动代替棋子的移动
当询问!的时候,判断每两个边界重合部分有多少点
当询问?的时候,如果点在边界上,则横纵坐标和边界相同,否则往棋盘相反的方向移动(不在边界上表示还未接触过棋盘的边界,因此每次移动都是K步不存在碰撞)
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #include<cmath> using namespace std; struct Point int x,y,flag; ; Point p[300005]; int n,m; int C[5]; int a[300005],b[300005]; int L,R,U,D,l,r,u,d,tl,tr,tu,td,DX,DY; int yin[10]; long long C2(long long a) return a*(a-1)/2; int main() yin[3]=1; yin[8]=2; yin[4]=3; yin[6]=4; int T; cin>>T; while(T--) cin>>n>>m; for(int i=0;i<=4;i++) C[i]=0; for(int i=1;i<=n;i++) scanf("%d%d",&p[i].x,&p[i].y); int t=p[i].x; p[i].x=p[i].y; p[i].y=n-t+1; p[i].flag=0; if(p[i].x==n) p[i].flag=1; else if(p[i].x==1) p[i].flag=2; if(p[i].y==1) C[yin[p[i].flag]*3]++; p[i].flag=3; else if(p[i].y==n) C[yin[p[i].flag*4]]++; p[i].flag=4; a[p[i].x]=i; b[p[i].y]=i; l=L=tl=1; r=R=tr=n; d=D=td=1; u=U=tu=n; DX=DY=0; char s[2]; int k; int flag1=0,flag2=0; while(m--) scanf("%s",s); if(s[0]==‘D‘) scanf("%d",&k); DY-=k; tu-=k; td-=k; tu=max(tu,1); td=max(td,1); if(D+k<=n) D+=k; U+=k; else D=n; U=2*n-1; if(flag1) continue; if(D>=u) flag1=1; continue; if(D>d) for(int i=d+1;i<=D;i++) int j=b[i]; C[yin[p[j].flag*3]]++; p[j].flag=3; d=D; else if(s[0]==‘U‘) scanf("%d",&k); DY+=k; tu+=k; td+=k; tu=min(tu,n); td=min(td,n); if(U-k>=1) U-=k; D-=k; else U=1; D=2-n; if(flag1) continue; if(U<=d) flag1=1; continue; if(U<u) for(int i=u-1;i>=U;i--) int j=b[i]; C[yin[p[j].flag*4]]++; p[j].flag=4; u=U; else if(s[0]==‘R‘) scanf("%d",&k); DX+=k; tl+=k; tr+=k; tl=min(tl,n); tr=min(tr,n); if(R-k>=1) R-=k; L-=k; else R=1; L=2-n; if(flag2) continue; if(R<=l) flag2=1; continue; if(R<r) for(int i=r-1;i>=R;i--) int j=a[i]; C[yin[p[j].flag*1]]++; p[j].flag=1; r=R; else if(s[0]==‘L‘) scanf("%d",&k); DX-=k; tl-=k; tr-=k; tl=max(tl,1); tr=max(tr,1); if(L+k<=n) L+=k; R+=k; else L=n; R=2*n-1; if(flag2) continue; if(L>=r) flag2=1; continue; if(L>l) for(int i=l+1;i<=L;i++) int j=a[i]; C[yin[p[j].flag*2]]++; p[j].flag=2; l=L; else if(s[0]==‘!‘) long long ans=0; if(flag1&&flag2) ans=C2(n); else if(flag1) ans=C2(l)+C2(n-r+1); else if(flag2) ans=C2(d)+C2(n-u+1); else ans=C2(C[1])+C2(C[2])+C2(C[3])+C2(C[4]); printf("%lld\n",ans); else if(s[0]==‘?‘) scanf("%d",&k); int ans1,ans2; if(p[k].x<=l||flag2) ans1=tl; else if(p[k].x>=r) ans1=tr; else ans1=p[k].x+DX; if(p[k].y<=d||flag1) ans2=td; else if(p[k].y>=u) ans2=tu; else ans2=p[k].y+DY; int t=ans2; ans2=ans1; ans1=n-t+1; printf("%d %d\n",ans1,ans2); return 0;
D.0:57:28(-1) solved by zcz
手推一下公式
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #include<cmath> using namespace std; double pi=3.14159265358979323; int main() int T; cin>>T; while(T--) double a,b,r,d; cin>>a>>b>>r>>d; double l=sqrt((a+r)*(a+r)+b*b); double ar=asin(b/l); d=d/180*pi; double ans; if(d>ar) ans=l-r; else ans=l*cos(ar-d)-r; printf("%.10lf\n",ans); return 0;
E.1:33:17 solved by gbs
容易想到要找到的数是前k个素数的乘积,且这个乘积<=n
根据样例可以找到规律,每多乘上一个素数p,答案乘上p/(p+1)
然后使用大数模板或者java,再gcd即可
对于每多乘上一个素数p,答案就乘上p/(p+1)的简略证明:
每多乘上一个素数p 并联的电阻的个数就变为原来的两倍 这些新的电阻的倒数是原来对应电阻倒数的1/p
那么所有电阻倒数和为原来倒数和的(1+p)/p 倒回来后为原来的p/(p+1)
#include <iostream> #include<stack> #include<math.h> #include<stdlib.h> #include<string.h> #include<string> #include<ctime> #include<complex> #include<stdio.h> #include<algorithm> #include<map> #include<queue> using namespace std; typedef unsigned long long LL; const int mod =1e9+7; const int maxn = 1e3+15; bool vis[maxn+15]; int prel[maxn/2+5]; int time1[maxn/2+5]; int pnum = 0; char an[105]; int bn[105]; int len1 ; void pre() memset(vis,false,sizeof(vis)); for (int i=2; i<maxn; i++) if (!vis[i]) //cout<<i<<endl; prel[pnum++] = i; for (int j =i<<1; j<maxn; j+=i) vis[j] = true; LL gcd(LL a,LL b) if (b == 0) return a; return gcd(b,a%b); void fen(LL n,int tp1) for (int i =0; i<pnum; i++) if (n==1) return ; while(n%prel[i] ==0) n/=prel[i]; time1[i]+=tp1; class bigint public: int bnum; int an[105]; bigint(int a = 0) memset(an,0,sizeof(an)); bnum = 1; an[0] = 0; if (a!= 0) bnum = 0; while(a) an[bnum] = a%10; a/=10; bnum++; void mul(int a) int jinwei = 0; for (int i=0; i<bnum; i++) an[i] = an[i]*a+jinwei; //123112312 jinwei = an[i]/10; an[i] =an[i]%10; if (jinwei >0) bnum = max(i+2,bnum); void out() for (int i=bnum-1; i>=0; i--) printf("%d",an[i]); ; bool op1(int pcd) int tuiwei = 0; bool if_ok = false;; //cout<<"@"<<pcd<<endl; for (int i=len1-1; i>=0; i--) bn[i] += tuiwei*10; tuiwei = bn[i]%pcd; bn[i] = bn[i]/pcd; //cout<<bn[i]; if (bn[i]>0) if_ok = true; //cout<<endl; return if_ok; int main() pre(); int t; cin >>t; while(t--) scanf("%s",an); len1 = strlen(an); for (int i=0; i<len1; i++) bn[i] = an[len1-1-i]-‘0‘; //cout<<bn[i]<<endl; memset(time1,0,sizeof(time1)); for (int i=0; i<pnum; i++) //cout<<"ce"<<endl; if (op1(prel[i])) //cout<<"we"<<i<<endl; fen(prel[i],1); fen(prel[i]+1,-1); else break; bigint fenzi(1),fenmu(1); for (int i=0; i<pnum; i++) while(time1[i]>0) fenzi.mul(prel[i]); --time1[i]; while(time1[i]<0) fenmu.mul(prel[i]); ++time1[i]; fenzi.out(); cout<<"/"; fenmu.out(); cout<<endl; return 0;
F.3:37:54(-3) solved by hl
WA原因:1.上下蜂巢可能会出现-- 或者 --这样的通路 2.编号标多了RE
建完图就是一个裸BFS,建图比较麻烦,建议分奇数行和偶数行,用每个六边形的中点标识整个六边形,
奇数行中点是(x,5) (x,17) (x,29) .....
偶数行中点是(x,11) (x,23) (x,35)......
然后每个六边形与其他六边形的边就是两个六边形中点连线的中点,判一下是不是空格就可以了,注意上下因为是---,有一个空格就算联通,需要特判
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read()int x = 0,f = 1;char c = getchar();while (c<‘0‘ || c>‘9‘)if (c == ‘-‘) f = -1;c = getchar(); while (c >= ‘0‘&&c <= ‘9‘)x = x * 10 + c - ‘0‘;c = getchar();return x*f; const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 3e6 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; const int a[6][2] = -1,-3,-1,3,1,-3,1,3,-2,0,2,0; map<PII,int>P; PII id[maxn]; LL dp[maxn]; char MAP[7010][7010]; int cnt,S,T; int main() int q; Sca(q); while(q--) int x,y; P.clear(); Sca2(x,y); cnt = 0; y = 6 * y + 3; x = 4 * x + 3; for(int i = 3; i < x ; i += 4) for(int j = 5; j <= y; j += 12) P[mp(i,j)] = ++cnt; id[cnt] = mp(i,j); for(int i = 5; i < x; i += 4) for(int j = 11; j <= y ; j += 12) P[mp(i,j)] = ++cnt; id[cnt] = mp(i,j); getchar(); for(int i = 1; i <= x; i ++) gets(MAP[i] + 1); for(int i = 1; i <= x; i ++) for(int j = 1; j <= y; j ++) if(MAP[i][j] == ‘S‘) S = P[mp(i,j)]; else if(MAP[i][j] == ‘T‘) T = P[mp(i,j)]; for(int i = 1; i <= cnt; i ++) int X = id[i].fi,Y = id[i].se; if(MAP[X - 2][Y - 1] == ‘ ‘ || MAP[X - 2][Y] == ‘ ‘ || MAP[X - 2][Y + 1] == ‘ ‘) MAP[X - 2][Y - 1] = MAP[X - 2][Y] = MAP[X - 2][Y + 1] = ‘ ‘; if(MAP[X + 2][Y - 1] == ‘ ‘ || MAP[X + 2][Y] == ‘ ‘ || MAP[X + 2][Y + 1] == ‘ ‘) MAP[X + 2][Y - 1] = MAP[X + 2][Y] = MAP[X + 2][Y + 1] = ‘ ‘; for(int i = 0; i <= cnt; i ++) dp[i] = INF; queue<int>Q; while(!Q.empty()) Q.pop(); Q.push(S); dp[S] = 1; while(!Q.empty() && dp[T] == INF) int u = Q.front(); Q.pop(); int X = id[u].fi,Y = id[u].se; for(int i = 0 ; i < 6; i ++) if(P[mp(X + a[i][0] * 2,Y + a[i][1] * 2)] && MAP[X + a[i][0]][Y + a[i][1]] == ‘ ‘) int v = P[mp(X + a[i][0] * 2,Y + a[i][1] * 2)]; if(dp[v] > dp[u] + 1) dp[v] = dp[u] + 1; Q.push(v); if(dp[T] == INF) dp[T] = -1; Prl(dp[T]); return 0;
I.0:25:19 solved by hl
策略画画就会发现是最左端取一个,最右端取一个,最左端取一个,最右端取一个......
pre表示当前左端取的数的下标和,cnt1表示左端取了多少数字
erp表示当前右边取的数的下表和,cnt2表示右端取了多少数字
一个数a[l]加入的贡献就是ans += a[l] * cnt1 - pre + (erp - a[l] * cnt2);
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read()int x = 0,f = 1;char c = getchar();while (c<‘0‘ || c>‘9‘)if (c == ‘-‘) f = -1;c = getchar(); while (c >= ‘0‘&&c <= ‘9‘)x = x * 10 + c - ‘0‘;c = getchar();return x*f; const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 1e5 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; LL pre,erp,a[maxn]; int main() int T = read(); while(T--) Sca(N); a[1] = 0; for(int i = 2; i <= N; i ++) Scl(a[i]); a[i] += a[i - 1]; LL ans = 0; int cnt = 1; int cnt1 = 0,cnt2 = 0; pre = erp = 0; int l = 1,r = N; for(int i = 1; i <= N ; i ++) if(cnt) ans += a[l] * cnt1 - pre + (erp - a[l] * cnt2); pre += a[l]; cnt1++; l++; else ans += a[r] * cnt1 - pre + (erp - a[r] * cnt2); erp += a[r]; cnt2++; r--; cnt ^= 1; printf("%lld",ans); if(i != N) printf(" "); puts(""); return 0;
H.unsolved by hl
应该是后缀数组加上某不知名数据结构去维护子区间的答案
某数据结构没有顶住
以上是关于2018ICPC 焦作 训练赛的主要内容,如果未能解决你的问题,请参考以下文章
[2018 ACM-ICPC 焦作赛区网络赛] G - Give Candies(找规律+快速幂)
[2018 ACM-ICPC 焦作赛区网络赛] H - String and Times(后缀自动机)
ICPC2018Beijing 现场赛D Frog and Portal 构造
03.28,周六,12:00-17:00,ICPC训练联盟周赛,选用试题:UCF Local Programming Contest 2016正式赛。