F.1:27:58(-3) solved by hl
打表发现所有满足条件的t可以通过f(n) = 4 * f(n - 1) - f(n - 2)递推出来
用__int128或者JAVA大数实现
#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 ", x) #define Prl(x) printf("%lld ",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 using namespace std; const int maxn = 2e6 + 10; __int128 a[maxn]; __int128 t = 1e32 + 10; __int128 read(){__int128 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;} void Outt(__int128 x){ if(!x) return; Outt(x / 10); putchar(x % 10 + '0'); } int cnt = 0; void init(){ a[1] = 4; a[2] = 14; for(cnt = 3; a[cnt - 1] < t; cnt++){ a[cnt] = 4 * a[cnt - 1] - a[cnt - 2]; } } int main(){ int T; Sca(T); init(); while(T--){ __int128 N = read(); int t = lower_bound(a + 1,a + 1 + cnt,N) - a; //if(a[t] != N) t++; Outt(a[t]); puts(""); } return 0; }
G.3:49:14(-1) solved by hl
答案的长度为N,假设答案有N层，每一层代表一位
设置一个栈表示当前所有可行的结点，显然可行结点的条件是结点表示的字符在这一层中最大。(贪心)
每次在上一层节点中找出可行的下一层结点递推
由于他的边是类似于随机构造的，出题人卡不掉我们
#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 ", x) #define Prl(x) printf("%lld ",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 using namespace std; const int maxn = 2e6 + 10; int N; int to[maxn]; struct Edge{ int to,next; }edge[maxn * 2]; int head[maxn],tot; void init(){ for(int i = 0 ; i <= N ; i ++) head[i] = -1; tot = 0; } void add(int u,int v){ edge[tot].to = v; edge[tot].next = head[u]; head[u] = tot++; } int Stack[2][maxn]; bool vis[maxn]; char str[maxn],ans[maxn]; int main(){ int T; Sca(T); int CASE = 1; while(T--){ Sca(N); scanf("%s",str); char Max = '0'; for(int i = 0 ; str[i]; i ++) Max = max(Max,str[i]); for(LL i = 0 ; i < N; i ++){ to[i] = (i * i + 1) % N; } int top[2]; top[0] = top[1] = 0; ans[1] = Max; for(int i = 0 ; i < N ; i ++){ if(str[i] == Max) Stack[1][++top[1]] = i; } for(int i = 2; i <= N; i ++){ char Max = '0'; for(int j = 1; j <= top[i + 1 & 1]; j ++){ int v = Stack[i + 1 & 1][j]; Max = max(Max,str[to[v]]); } ans[i] = Max; for(int j = 1; j <= top[i + 1 & 1]; j ++){ int v = Stack[i + 1 & 1][j]; if(str[to[v]] < Max) continue; if(vis[to[v]]) continue; vis[to[v]] = 1; Stack[i & 1][++top[i & 1]] = to[v]; } for(int j = 1; j <= top[i & 1]; j ++) vis[Stack[i & 1][j]] = 0; top[i + 1 & 1] = 0; } printf("Case #%d: ",CASE++); for(int i = 1 ; i <= N; i ++) printf("%c",ans[i]); puts(""); } return 0; } /* 4 3 149 5 12345 1 7 3214567 9 261025520 */
I.0:30:55(-1) solved by hl
4个2 ^ 62相加会爆ULL
特判一下或者直接上大数
#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 ", x) #define Prl(x) printf("%lld ",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 using namespace std; void Outt(__int128 x){ if(!x) return; Outt(x / 10); putchar(x % 10 + '0'); } int main() { int t; cin >> t; LL a,b,c,d; while(t--) { cin >> a >> b >> c >> d; __int128 ans = 0; ans += a; ans += b; ans += c; ans += d; if(ans){ Outt(ans); puts(""); } else puts("0"); } return 0; }
K.0:26:27 solved by zcz
推公式
#include <iostream> #include<cstring> #include<cstdio> using namespace std; int a[505]; int main() { int T; cin>>T; int n; while(T--) { cin>>n; for(int i=1;i<=n;i++) { scanf("%d",&a[i]); } printf("%d ",a[n]-a[1]-n+1-min(a[2]-a[1]-1,a[n]-a[n-1]-1)); } return 0; }
L.1:10:36 solved by hl
一条边成立（计入贡献）的条件是这条边两端点形成的子树size >= K
dfs求一个子树和就可以了
#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 ", x) #define Prl(x) printf("%lld ",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 using namespace std; const int maxn = 2e5 + 10; struct Edge{ int to,next; }edge[maxn * 2]; int head[maxn],tot; int N,K; void init(){ for(int i = 0 ; i <= N ; i ++) head[i] = -1; tot = 0; } void add(int u,int v){ edge[tot].to = v; edge[tot].next = head[u]; head[u] = tot++; } int sz[maxn]; void dfs(int t,int la){ sz[t] = 1; for(int i = head[t]; ~i ; i = edge[i].next){ int v = edge[i].to; if(v == la) continue; dfs(v,t); sz[t] += sz[v]; } } int ans = 0; void dfs2(int u,int la){ for(int i = head[u]; ~i ; i = edge[i].next){ int v = edge[i].to; if(v == la) continue; if(sz[v] >= K && N - sz[v] >= K) ans++; dfs2(v,u); } } int main(){ int T; Sca(T); while(T--){ Sca2(N,K); init(); for(int i = 1; i <= N - 1; i ++){ int u,v; Sca2(u,v); add(u,v); add(v,u); } dfs(1,-1); ans = 0; dfs2(1,-1); Pri(ans); } return 0; }
M.4:21:33 solved by zcz and hl
设置每个点的权值为他上下左右可以移动的方向 + 它本身
即一个点初始为5,四周边每有一个障碍物 - 1，它本身就是障碍物为0
最终答案就是右下角三角形的 (N * (N - 1) / 2)个点的权值和 / 所有点的权值和
#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 ", x) #define Prl(x) printf("%lld ",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 using namespace std; const int maxn = 2e6 + 10; LL N,K; map<PII,int>Q; map<PII,bool> vis; int id(int x,int y){ int sum = 5; if(x == N || x == 1) sum--; if(y == N || y == 1) sum--; sum -= Q[mp(x,y)]; return sum; } int main(){ int T; Sca(T); int CASE = 1; while(T--){ Scl(N); Scl(K); Q.clear(); vis.clear(); LL sum = N * (N + 1) / 2; LL up = 0; up = (N - 2) * 4 * 2 + 9; up += (sum - N - (N - 1)) * 5; LL down = (N - 2) * 4 * 4 + 4 * 3 + (N - 2) * (N - 2) * 5; LL dd = 0,du = 0; for(int i = 1; i <= K ; i ++){ int x,y; Sca2(x,y); x++; y++; if(x != 1 && !vis[mp(x - 1,y)]){ if(x - 1 + y >= N + 1) du++; dd++; Q[mp(x - 1,y)]++; } if(x != N && !vis[mp(x + 1,y)]){ if(x + 1 + y >= N + 1) du++; dd++; Q[mp(x + 1,y)]++; } if(y != 1 && !vis[mp(x,y - 1)]){ if(x - 1 + y >= N + 1) du++; dd++; Q[mp(x,y - 1)]++; } if(y != N && !vis[mp(x,y + 1)]){ if(x + 1 + y >= N + 1) du++; dd++; Q[mp(x,y + 1)]++; } vis[mp(x,y)] = 1; if(x + y >= N + 1) du += id(x,y); dd += id(x,y); } up -= du; down -= dd; LL g = __gcd(up,down); up /= g; down /= g; printf("Case #%d: %lld/%lld ",CASE++,up,down); } return 0; } /* 4 3 149 5 12345 1 7 3214567 9 261025520 */