2018CCPC吉林 训练赛题解
A.
00:09:13
solved by hl
打个表发现规律是3个odd,5个even,7个odd,9个even.......
很显然是个等差数列，二分判断项数的奇偶即可
不过似乎大家都有更加简单的方法
#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; LL check(LL x){ return x * (2 + x); } void solve(LL x){ LL l = 1,r = x; LL ans = 0; while(l <= r){ LL m = l + r >> 1; if(check(m) >= x){ ans = m; r = m - 1; }else{ l = m + 1; } } if(ans & 1) puts("odd"); else puts("even"); } int main(){ int T = read(); int CASE = 1; while(T--){ LL t = read(); printf("Case %d: ",CASE++); solve(t); } return 0; }
B.
0:26:03
solved by gbs
没有什么好讲的，只有一些细节需要注意。观察可知，当hour为0时,hour输出为12方可符合题目的格式。另外分钟数不足10时前面需要添1个0
#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> using namespace std; typedef long long LL; void outhour(int a) { if (a==0) cout<<12; else cout<<a; } void out(int a) { if (a<10)cout<<"0"; cout<<a; } int main() { int t; cin >>t; int t1 =1; int hour,minute; char lls; string apm; string city1; string days; map<string,int>map1; map1["Beijing"] = 8; map1["Washington"] = -5; map1["London"] = 0; map1["Moscow"] = 3; while(t--) { cin >>hour>>lls>>minute; if (hour == 12) hour-=12; cin >>apm; if (apm == "PM") hour+=12; cin>>city1; hour-=map1[city1]; cin>>city1; hour+=map1[city1]; if (hour <0) { hour+=24; days = "Yesterday"; } else if (hour>=24) { hour-=24; days = "Tomorrow"; } else days = "Today"; if (hour >= 12) { hour -= 12; apm = "PM"; } else apm ="AM"; printf("Case %d: ",t1++); cout<<days<<' '; outhour(hour); cout<<":"; out(minute); cout<<" "; cout<<apm; cout<<endl; } return 0; }
C.
0:43:44
solved by hl
将所有元素从大到小排序，按照两个相同的k并为一个k - 1的方式。
形如哈夫曼树一样维护一个优先队列不断安排上，如果最终可以并为至少两个1存在，就可行，其中一个1的元素全为0组，另一个全为1组
至于怎么确定哪个元素是哪个组的，可以并查集
#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; PII a[maxn]; struct node{ int id,k; node(){} node(int id,int k):id(id),k(k){} friend bool operator < (node a,node b){ return a.k < b.k; } }; int fa[maxn]; void init(){ for(int i = 1; i <= N; i ++) fa[i] = i; } int find(int x){ if(x == fa[x] || x < 0) return x; return fa[x] = find(fa[x]); } int main(){ int T = read(); int CASE = 1; while(T--){ Sca(N); init(); priority_queue<node>Q; for(int i = 1; i <= N ; i ++){ Sca(a[i].fi); a[i].se = i; Q.push(node(a[i].se,a[i].fi)); } bool flag = 0; while(!Q.empty()){ node u = Q.top(); Q.pop(); if(!Q.empty() && u.k == Q.top().k){ node v = Q.top(); Q.pop(); if(u.k == 1){ flag = 1; fa[u.id] = -1; fa[v.id] = -2; break; } fa[u.id] = v.id; v.k--; // cout << v.k <<endl; Q.push(v); } } printf("Case %d: ",CASE++); if(!flag){ puts("NO"); continue; } puts("YES"); for(int i = 1; i <= N ; i ++){ int x = find(i); if(x == i || x == -1) printf("1"); else printf("0"); } puts(""); } return 0; }
D.
0:57:47
solved by gbs
dp
开一个大小为两百的数组dp[205],下标为i,表示进行若干局游戏后，当前的(q*2)是i的概率。对于每一个i,进行一局有 p/100*i*2/200的概率结束游戏，把答案加上对应的期望，另外有（1-p/100）的概率让i+3,其余的情况下i+4（i最大为200）。根据当前的情况可以推出下一局游戏时的情况。
若干局后只有dp[200]不为0，那么只要一次胜利，即可成功，期望是前面期望的累计再加上100/q
#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> using namespace std; typedef long long LL; double dp[205]; int main() { int t; int t1= 1; int p1; cin >>t; while(t--) { cin >>p1; for (int i=0; i<205; i++) dp[i] = 0; dp[4] = 1; double win_rate; double over_rate; double fin_ans = 0; for (int times=1;; times++) { int min_poi =-1; for (int i=200; i>=0; i--) { if (dp[i] != 0) { min_poi = i; win_rate = dp[i] *p1/100; over_rate = win_rate *i/200; dp[i] -= over_rate; fin_ans += times*over_rate; if (i!= 200) { dp[min(200,i+4)] += win_rate-over_rate; dp[min(200,i+3)] += dp[i] - win_rate + over_rate; dp[i] = 0; } } } if (min_poi == -1)break; if (min_poi == 200) { fin_ans += dp[200] *(times + 100.0/p1); break; } } printf("Case %d: %.9f\n",t1++,fin_ans); } return 0; }
E.
4:07:54
solved by gbs
正解可能是要推公式，但是用三分也可以做。
对于任意的时间，若0<z<h 那么有令f(x) = 当前截面的圆的半径的平方 - 当前时间下点到圆锥轴线的距离的平方 若f(x)>=0则点在圆锥内 否则 点在圆锥外
由题意知 点先在圆锥外，后到圆锥内，再然后可能到圆锥外 那么f(x)先<0 后>0, 再然可能<0 第一个f(x) =0的点就是答案
由于f(x)是两个二次或以下的函数相加，f(x)也必然是二次或以下的函数
如果f(x)是二次函数，那么用三分可以求出最大值。在时间的下限和这个最大值对应的时间中二分找到f(x)=0的时间即是答案
如果f(x)是一次函数，直接套用三分也可成功（此时三分的上限就是二分的上限）
#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<deque> using namespace std; typedef long long LL; const double dis = 1e-9; double r,h; double cx0,cy0,cz0; double vx,vy,vz; double llx,lly,llz,lltime; void find_xyz(double timea) { lltime = timea; llz = cz0 + vz*lltime; llx = cx0 + vx*lltime; lly = cy0 + vy*lltime; } inline double find_r2(double z) { if (h-z<-dis)return 0; if (z<-dis)return 0; double llr = (h-z)*r/h; return llr*llr; } double jval(double time1) { find_xyz(time1); return find_r2(llz)-llx*llx-lly*lly; } int main() { int t; int case1 = 1; cin >>t; while(t--) { cin >>r>>h; cin >>cx0>>cy0>>cz0; cin >>vx>>vy>>vz; double time1 =0 ; double time2 = 10000; double the_mid; if (vz >0) { time2 = min(time2,(h-cz0)/vz+ 2*dis); time2 = max(time2,time1 + 2*dis); } else if (vz <0) { time2 = min(time2,(0-cz0)/vz+ 2*dis); time2 = max(time2,time1 + 2*dis); if (cz0 >h) { time1 = max(time1,(h-cz0)/vz- 2*dis); time1 = min(time1,time2 - 2*dis); } } double t1 =time1; double t2 = time2; double mid; double right1; double mval; double rval; while (t2-t1>dis) { mid =(t2+t1)/2; right1 = (mid + t2)/2; mval = jval(mid); rval = jval(right1); if (mval >rval) t2 = right1; else t1 = mid; } the_mid = t2; while(the_mid - time1 >dis) { mid =(time1+the_mid)/2; mval = jval(mid); if (mval <0) time1 = mid; else the_mid = mid; } printf("Case %d: %.10f\n",case1++,time1); } return 0; }
F.
1:32:14
solved by gbs
点i-1必然不能完美收到点i的信号（因为找不到j），但对于点k=i-2以及它左边的点（只要能收到i的信号），只要令j=i-1即可让能收到信号的点k完美收到信号。
说明：
k点可以完美收到i的信号的条件如下：
k<i,
k能收到i的信号，
且存在k<j<i,使 j到k距离大于等于j到i距离
且k和i都能收到j的信号、
由题目条件可知，当位于i的点能向左传到k点时，点i-1必然可以向左传到k点时。i和i-1的距离是1，也必然满足距离的条件。因为点i-1能传到k，那么rad至少为2，那么点i-1也可以传到i点。那么所有i能传到且与i距离大于等于2的点都符合条件。
所以对于单个点，答案为max(0,rad-2) 只要把所有答案异或出来即可
#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 = 1e6 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; int a[maxn]; int main(){ int T = read(); int CASE = 1; while(T--){ Sca(N); for(int i = 1; i <= N ; i ++) Sca(a[i]); int l = 0; LL ans = 0; for(int r = 1; r <= N ; r ++){ if(a[r] >= 2) ans ^= a[r] - 2; } printf("Case %d: ",CASE++); Prl(ans); } return 0; }
H.
04:37:22 （-2）
solved by hl
线段树维护区间和，主要难点在update
每个节点维护一个区间和sum以及一个区间位数和dig
打三个lazy标记，一个维护左边增加的数L，一个维护右边增加的数R，一个维护左右两边增加的数的位数lazydig。
其中位数描述成十的倍数，1位为10,2位为100,3位为1000
那么更新区间和就是sum = sum * dig + R * len + L * dig * lazydig
其中len为区间长度，除了sum之外其他的比较好维护
#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; struct Tree{ int l,r; LL dig,sum,lazydig,L,R; }tree[maxn << 2]; LL ten[maxn]; inline LL mul(LL a,LL b){ return a % mod * b % mod; } inline LL add(LL a,LL b){ return ((a + b) % mod + mod) % mod; } void Pushup(int t){ tree[t].sum = add(tree[t << 1].sum,tree[t << 1 | 1].sum); tree[t].dig = add(tree[t << 1].dig,tree[t << 1 | 1].dig); } void Build(int t,int l,int r){ tree[t].l = l; tree[t].r = r; tree[t].sum = 0; tree[t].dig = tree[t].lazydig = 1; tree[t].L = tree[t].R = 0; if(l == r) return; int m = l + r >> 1; Build(t << 1,l,m); Build(t << 1 | 1,m + 1,r); Pushup(t); } void change(int t,LL L,LL R,LL dig){ int len = tree[t].r - tree[t].l + 1; tree[t].sum = add(add(mul(tree[t].sum,dig),mul(R,len)),mul(mul(L,dig),tree[t].dig)); tree[t].dig = mul(tree[t].dig,mul(dig,dig)); tree[t].L = add(mul(tree[t].lazydig,L),tree[t].L); tree[t].R = add(R,mul(tree[t].R,dig)); tree[t].lazydig = mul(tree[t].lazydig,dig); } void Pushdown(int t){ if(tree[t].lazydig != 1){ change(t << 1,tree[t].L,tree[t].R,tree[t].lazydig); change(t << 1 | 1,tree[t].L,tree[t].R,tree[t].lazydig); tree[t].L = tree[t].R = 0; tree[t].lazydig = 1; } } void update(int t,int l,int r,LL d){ if(l <= tree[t].l && tree[t].r <= r){ change(t,d,d,10); return; } Pushdown(t); int m = (tree[t].l + tree[t].r) >> 1; if(r <= m) update(t << 1,l,r,d); else if(l > m) update(t << 1 | 1,l,r,d); else{ update(t << 1,l,m,d); update(t <<1 | 1,m + 1,r,d); } Pushup(t); } LL query(int t,int l,int r){ if(l <= tree[t].l && tree[t].r <= r) return tree[t].sum; Pushdown(t); int m = (tree[t].l + tree[t].r) >>1; if(r <= m) return query(t << 1,l,r); else if(l > m) return query(t << 1 | 1,l,r); else{ return add(query(t << 1,l,m),query(t << 1 | 1,m + 1,r)); } } int main(){ int T = read(); ten[0] = 1; int CASE = 1; for(int i = 1 ; i < maxn; i ++) ten[i] = ten[i - 1] * 10 % mod; while(T--){ Sca2(N,M); Build(1,1,N); printf("Case %d:\n",CASE++); while(M--){ char op[10];int l,r; scanf("%s%d%d",op,&l,&r); if(op[0] == 'w'){ int d = read(); update(1,l,r,d); }else{ Prl(query(1,l,r)); } } } return 0; }
I.
1:57:45
solved by gbs
贪心
分为2种方案，一种把对面的怪打完，剩余的怪直接攻击。一种不把对面的怪打完。
对于第一种情况：按攻击力降序枚举b所拥有的怪物，每次把所有攻击力大于对面的己方怪物加入双端队列。如果对面是防御表示，则派攻击力最低的去攻击，否则派攻击力最高的去攻击。若找不到怪则失败。
第二种情况：
首先排除所有防御表示的怪。自己每一次攻击，收益是自己的攻击力-对方攻击力（忽略攻击不足的情况）。若选择攻击k轮，那么收益是自己前k大-对方前k小。所以只需让自己最强的怪攻击对方最弱的怪，直到不能攻击为止。
#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<deque> using namespace std; typedef long long LL; int n,m; int an[100005]; int bn[100005]; int cn[100005]; LL sum_atk; LL fin_ans; void judge1() { deque<int>stk1; int now_p = n-1; LL llans =sum_atk; for (int i=m-1; i>=0; i--) { while(now_p>=0) { if (an[now_p]>bn[i]) { stk1.push_back(an[now_p]); now_p--; } else { break; } } if (stk1.empty()) return ; if (cn[i] == 1) { llans -= stk1.back(); stk1.pop_back(); } else { llans -= bn[i]; stk1.pop_front(); } } fin_ans = max(fin_ans,llans); } void judge2() { int now_n =n-1; int now_m =0; LL llans = 0; while(now_m<m &&now_n>=0) { if (cn[now_m] ==1) { now_m++; continue; } if (an[now_n] > bn[now_m]) { llans += an[now_n] - bn[now_m]; now_n--; now_m++; } else break; } fin_ans = max(fin_ans,llans); } int main() { int t; int t1 = 1; cin >>t; while(t--) { cin >>n >>m; sum_atk = 0; for (int i=0; i<n; i++) { scanf("%d",&an[i]); sum_atk += an[i]; } for (int i=0; i<m; i++) scanf("%d",&bn[i]); for (int i=0; i<m; i++) scanf("%d",&cn[i]); sort(an,an+n); sort(bn,bn+m); fin_ans = 0; judge1(); //cout<<fin_ans<<endl; judge2(); printf("Case %d: %lld\n",t1++,fin_ans); } return 0; }
