Codeforces Round #426 (Div. 1) (ABCDE)
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1. 833A The Meaningless Game
大意: 初始分数为$1$, 每轮选一个$k$, 赢的人乘$k^2$, 输的人乘$k$, 给定最终分数, 求判断是否成立.
判断一下$acdot b$是否是立方数, 以及能否被那个立方的因子整除即可. cbrt竟然有误差, 特判了一下, 好坑
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ‘ ‘ #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<‘,‘;hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<‘0‘||p>‘9‘)p=getchar();while(p>=‘0‘&&p<=‘9‘)x=x*10+p-‘0‘,p=getchar();return x;} //head int main() { int n; scanf("%d", &n); while (n--) { int a, b; scanf("%d%d", &a, &b); ll x = (ll)a*b, t = cbrt(x); while (t*t*t>x) --t; while (t*t*t<x) ++t; puts(t*t*t==x&&a%t==0&&b%t==0?"Yes":"No"); } }
2. 833B The Bakery
大意: 给定$n$元素序列, 求划分为最多$x$段, 每段的贡献为不同元素数, 求最大贡献.
设$dp_{i,j}$为前$i$个数分成$j$段的最大值, 开$k$棵线段树, 位置$x$维护dp[x-1][j]+[x...i]的贡献即可
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ‘ ‘ #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<‘,‘;hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<‘0‘||p>‘9‘)p=getchar();while(p>=‘0‘&&p<=‘9‘)x=x*10+p-‘0‘,p=getchar();return x;} //head const int N = 35100; int n,k,a[N],pre[N],vis[N]; int dp[N][50]; //dp[i][j] = 前i个数,分成j段的最大值 //要支持区间加, 查询区间最值 struct { int ma[N<<2],tag[N<<2]; void pd(int o) { if (tag[o]) { ma[lc]+=tag[o],tag[lc]+=tag[o]; ma[rc]+=tag[o],tag[rc]+=tag[o]; tag[o]=0; } } void add(int o, int l, int r, int ql, int qr, int v) { if (ql<=l&&r<=qr) return ma[o]+=v,tag[o]+=v,void(); pd(o); if (mid>=ql) add(ls,ql,qr,v); if (mid<qr) add(rs,ql,qr,v); ma[o]=max(ma[lc],ma[rc]); } } tr[55]; int main() { scanf("%d%d", &n, &k); REP(i,1,n) { scanf("%d",a+i); pre[i] = vis[a[i]]; vis[a[i]] = i; } REP(i,1,n) { REP(j,0,k-1) tr[j].add(1,1,n,pre[i]+1,i,1); REP(j,1,k) dp[i][j] = tr[j-1].ma[1]; if (i==n) break; REP(j,1,k-1) tr[j].add(1,1,n,i+1,i+1,dp[i][j]); } printf("%d ", dp[n][k]); }
3. 833C Ever-Hungry Krakozyabra
大意: 给定$L,R$, 将区间$[L,R]$的每个数去除数位$0$后对数位进行排序, 求一共能得到多少种数.
先特判掉1e18, 所有可能得到的数最多为$inom{18+9}{9}=1562275$, 可以直接暴力枚举每个数$x$, 判断$x$是否能在区间$[L,R]$得到即可. 判断合法性本来写的$O(18^2)$的暴力贪心, 结果TLE on test 70. 按照官方题解改成复杂度$O(10*18)$的$dfs$就过了...... 看评论区说似乎有不用暴力枚举的做法, 没太懂
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ‘ ‘ #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<‘,‘;hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<‘0‘||p>‘9‘)p=getchar();while(p>=‘0‘&&p<=‘9‘)x=x*10+p-‘0‘,p=getchar();return x;} //head int ans,len,flag,cnt,a[20],x[20],y[20]; ll L, R, fac[20]; int solve(int d, int l1, int l2) { if (!l1&&!l2||d<0) return 1; int U=l1?x[d]:0,D=l2?y[d]:9; REP(i,U,D) if (a[i]) { --a[i]; if (solve(d-1,l1&&x[d]==i,l2&&y[d]==i)) return ++a[i],1; ++a[i]; } return 0; } int chk() { return solve(len-1,1,1); } //生成所有合法的数 void dfs(int d, int pre) { if (d>len) return; REP(i,pre,9) { ++a[i],++cnt; a[0] = len-cnt; ans += chk(); dfs(d+1,i); --a[i],--cnt; } } int main() { fac[0] = 1; REP(i,1,18) fac[i] = fac[i-1]*10; scanf("%lld%lld",&L,&R); if (R==fac[18]) --R, flag = 1; if (L==fac[18]) --L; ll t = L; while (t) x[len++]=t%10,t/=10; t = R, len = 0; while (t) y[len++]=t%10,t/=10; dfs(1,1); printf("%d ",ans); }
4.
5. 833E Caramel Clouds
大意: 给定$n$朵云, 第$i$朵出现时间范围$[l_i,r_i]$, 删除需要花费$c_i$, 初始时刻为$0$, 预算为$C$. 给定$m$个询问$k_i$, 求一棵需要照$k_i$时间阳光的花最快什么时间长成, 最多删除两朵云, 询问独立.
参考的这个https://www.cnblogs.com/ECJTUACM-873284962/p/7265224.html
主要思路就是依次处理每个时间段, 预处理每个时刻的最大光照时间.
#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <map> #include <queue> #include <string> #include <cstring> #include <bitset> #include <functional> #include <random> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ‘ ‘ #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<‘,‘;hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<‘0‘||p>‘9‘)p=getchar();while(p>=‘0‘&&p<=‘9‘)x=x*10+p-‘0‘,p=getchar();return x;} //head #ifdef ONLINE_JUDGE const int N = 1e6+50; #else const int N = 1e2+10; #endif int n, C, cost[N], tag[N], b[N]; pii tr1[N],tr2[N]; //树状数组维护区间最大次大 void add(int id, int x, int v) { x = lower_bound(b+1,b+1+*b,x)-b; for (; x<=*b; x+=x&-x) { if (tr1[x].y==id) tr1[x].x = max(tr1[x].x,v); else if (tr2[x].x<v) tr2[x] = pii(v,id); if (tr1[x]<tr2[x]) swap(tr1[x],tr2[x]); } } int qry(int id, int x) { x = upper_bound(b+1,b+1+*b,x)-b-1; int ret = 0; for (; x; x^=x&-x) { if (id==tr1[x].y) ret=max(ret,tr2[x].x); else ret=max(ret,tr1[x].x); } return ret; } int main() { scanf("%d%d", &n, &C); vector<pii> events; REP(i,1,n) { int l,r; scanf("%d%d%d",&l,&r,cost+i); events.pb(pii(l,i)); events.pb(pii(r,i)); b[i] = cost[i]; } events.pb(pii(0,0)); events.pb(pii(2e9,0)); sort(events.begin(),events.end()); sort(b+1,b+1+n),*b=unique(b+1,b+1+n)-b-1; REP(i,0,*b+5) tr1[i].y=tr2[i].y=-1; vector<pii> ans; map<pii,int> len; set<int> s; int mx = 0; if (events[0].y) s.insert(events[0].y); for (int i=1; i<events.size(); ++i) { int d = events[i].x-events[i-1].x; if (d>0&&s.size()<=2) { int p = 0, q = 0; if (s.size()) p = *s.begin(); if (s.size()==2) q = *s.rbegin(); int start = -1; if (!p) { start = mx; len[{0,0}] += d; } else if (!q) { if (cost[p]<=C) { start = len[{p,0}]+len[{0,0}]; //找一个q, 使得len[{p,q}]+len[{q,0}]的最大 int ma = tag[p]; //相交的情况 //不相交的情况 ma = max(ma, qry(p,C-cost[p])); start += ma; len[{p,0}] += d; add(p,cost[p],len[{p,0}]); } } else if (cost[p]+cost[q]<=C) { start = len[{p,q}]+len[{p,0}]+len[{q,0}]+len[{0,0}]; len[{p,q}] += d; tag[p] = max(tag[p], len[{p,q}]+len[{q,0}]); tag[q] = max(tag[q], len[{p,q}]+len[{p,0}]); } if (~start&&start+d>mx) { mx = start+d; ans.pb(pii(mx, events[i].x)); } } int id = events[i].y; if (!id) continue; if (s.count(id)) s.erase(id); else s.insert(id); } int m; scanf("%d", &m); while (m--) { int k; scanf("%d", &k); auto p = lower_bound(ans.begin(),ans.end(),pii(k,0)); printf("%d ", p->y-(p->x-k)); } }
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