ACM-ICPC 2018 徐州赛区网络预赛
Posted solvit
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ACM-ICPC 2018 徐州赛区网络预赛相关的知识，希望对你有一定的参考价值。
#include <bits/stdc++.h> using namespace std; #define N 1000005 long long mod = 1e9 + 7; long long power(long long a,long long b){ a %= mod; long long ret = 1; while(b){ if(b & 1) ret = ret * a % mod; a = a * a % mod; b >>= 1; } return ret; } long long re[N],inv[N],fac[N]; void init(int n){ re[0] = inv[1] = fac[0] = 1; for(int i = 1;i <= n;++i) fac[i] = fac[i-1] * i % mod; for(int i = 2;i <= n;++i) inv[i] = (mod-mod/i)*inv[mod%i] % mod; for(int i = 1;i <= n;++i) re[i] = re[i-1] * inv[i] % mod; } long long c(int a,int b){ if(a < b) return 0; return fac[a]*re[b]%mod*re[a-b]%mod; } int main() { init(1000000); int T; cin >> T; while(T--){ long long k,n; scanf("%lld%lld",&n,&k); long long ans = 0; long long tot = power(2LL,k); long long pre = 1; for(int i = n-1;i >= 0;--i){ pre = pre * tot % mod; ans += 1LL * ((i&1)?-1:1) * c(n,i) * pre % mod; ans = (ans + mod) % mod; } printf("%lld ",(ans+(n&1?0:1)*tot)%mod); } return 0; }
#include <bits/stdc++.h> using namespace std; int sta[210][1005]; int a[1005],b[1005],c[1005]; int main() { int n,m,l,r; scanf("%d%d%d%d",&n,&m,&r,&l); for(int i = 1;i <= n;++i){ scanf("%d%d%d",&a[i],&b[i],&c[i]); } for(int i = -100;i <= 100;++i){ if(i <= l) sta[i+100][n+1] = 1;//第二个人赢的位置 else if(i <= r) sta[i+100][n+1] = 2; else sta[i+100][n+1] = 3;//第一个人赢 } for(int i = n;i >= 1;--i){ for(int j = -100;j <= 100;++j){ if(i & 1){ sta[j+100][i] = 1; //第一个人操作 if(a[i]) sta[j+100][i] = max(sta[min(j+100+a[i],200)][i+1],sta[j+100][i]); if(b[i]) sta[j+100][i] = max(sta[max(0,j+100-b[i])][i+1],sta[j+100][i]); if(c[i]) sta[j+100][i] = max(sta[-j+100][i+1],sta[j+100][i]); } else{ sta[j+100][i] = 3; if(a[i]) sta[j+100][i] = min(sta[min(j+100+a[i],200)][i+1],sta[j+100][i]); if(b[i]) sta[j+100][i] = min(sta[max(0,j+100-b[i])][i+1],sta[j+100][i]); if(c[i]) sta[j+100][i] = min(sta[-j+100][i+1],sta[j+100][i]); } } } if(sta[m+100][1] == 1) puts("Bad Ending"); else if(sta[m+100][1] == 2) puts("Normal Ending"); else puts("Good Ending"); return 0; }
#include<bits/stdc++.h> using namespace std; typedef long long ll; map<pair<int,int>,int> M; map<pair<int,int>,int> vis; int t,n,k,a,b; int ans; int main() { scanf("%d",&t); while(t--) { scanf("%d",&n); vis.clear();M.clear(); ans = 1; for(int i=1;i<=n;i++) { scanf("%d",&k); for(int j=0;j<k;j++){ scanf("%d%d",&a,&b); if(vis[make_pair(a,b)] == i-1){ M[make_pair(a,b)]++; ans = max(M[make_pair(a,b)],ans); } else if(vis[make_pair(a,b)] == i){ continue; } else{ M[make_pair(a,b)] = 1; } vis[make_pair(a,b)] = i; } } cout<<ans<<endl; } return 0; }
#include<bits/stdc++.h> using namespace std; typedef long long ll; struct point{ int x,y; bool operator < (const point & _point)const{ return x < _point.x; } }tot[50005]; set<point> M; int n,a,b; int main() { scanf("%d",&n); for(int i=1;i<=n;i++) { scanf("%d%d",&tot[i].x,&tot[i].y); } M.clear(); ll ans = 0; ans += tot[n].x;ans += tot[n].y; M.insert(tot[n]); //cout<<ans<<endl; for(int i=n-1;i>=1;i--) { if(M.lower_bound(tot[i])==M.end()){ set<point>::iterator it = M.end(); it--; point temp = *(it); ans += tot[i].y; ans += tot[i].x - temp.x; // cout<<"11111111 "<<temp.x<<" "<<temp.y<<endl; } else if(M.lower_bound(tot[i])==M.begin()){ point temp = *M.begin(); ans += tot[i].x; ans += tot[i].y - temp.y; //cout<<"22222222 "<<temp.x<<" "<<temp.y<<endl; } else{ set<point>::iterator it = (M.lower_bound(tot[i])); point temp1 = *it; it--; point temp = *it; ans += tot[i].y - temp1.y; ans += tot[i].x - temp.x; //cout<<"13333331 "<<temp.x<<" "<<temp.y<<endl; // cout<<"14444444 "<<temp1.x<<" "<<temp1.y<<endl; } M.insert(tot[i]); // cout<<ans<<endl; } cout<<ans<<endl; }
#include <bits/stdc++.h> using namespace std; #define N 100005 #define lc (d<<1) #define rc (d<<1|1) #define mid ((l+r)>>1) long long v[N]; struct Tr{ long long sum,dp,cnt; Tr operator + (const Tr A) const{ Tr B; B.cnt = A.cnt + cnt; B.sum = A.sum + sum; B.dp = A.dp + sum * A.cnt + dp; return B; } }tr[N<<2]; void build(int l,int r,int d){ if(l == r){ tr[d].cnt = 1; tr[d].dp = tr[d].sum = v[l]; return; } build(l,mid,lc),build(mid+1,r,rc); tr[d] = tr[lc] + tr[rc]; } void update(int l,int r,int pos,int val,int d){ if(l == r){ tr[d].sum = tr[d].dp = val; return; } if(pos <= mid) update(l,mid,pos,val,lc); else update(mid+1,r,pos,val,rc); tr[d] = tr[lc] + tr[rc]; } Tr query(int l,int r,int L,int R,int d){ if(L == l && R == r) return tr[d]; if(R <= mid) return query(l,mid,L,R,lc); if(L > mid) return query(mid+1,r,L,R,rc); return query(l,mid,L,mid,lc) + query(mid+1,r,mid+1,R,rc); } int main() { int n,q; cin >> n >> q; for(int i = 1;i <= n;++i) scanf("%lld",&v[i]); build(1,n,1); int op,L,R; for(int i = 1;i <= q;++i){ scanf("%d%d%d",&op,&L,&R); if(op == 1) printf("%lld ",query(1,n,L,R,1).dp); else update(1,n,L,R,1); } return 0; }
#include <bits/stdc++.h> using namespace std; int T,n; char s[1000005],z; int main(){ scanf("%d",&T); while(T--){ scanf("%d %c",&n,&z); scanf("%s",s); int ans = strlen(s) * 2; for (int i = 0;i < strlen(s);++i){ int tmp = abs(s[i] - z); //cout << tmp << endl; if (tmp == 0) ans -= 2; else if (tmp < 10 && tmp > 0){ ans--; break; }else if (tmp >= 10) break; } if (ans == 0) puts("1"); else printf("%d ",ans); } }
以上是关于ACM-ICPC 2018 徐州赛区网络预赛的主要内容，如果未能解决你的问题，请参考以下文章
ACM-ICPC 2018 徐州赛区网络预赛 D. EasyMath
ACM-ICPC 2018 徐州赛区网络预赛 J Maze Designer
ACM-ICPC 2018 徐州赛区网络预赛 J. Maze Designer (最大生成树+LCA求节点距离)