2019 ICPC 上海 网络赛
摸了一整场的鱼，签了个到，做了个FFT还忘记初始化赛后才发现
B.00:16:52 solved by hl
很显然前缀和搞搞就行了，但是卡O(N),要O(M)
#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 = 1e4 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; map<int,int>pre; PII q[maxn]; int main(){ int T = read(); int Case = 1; while(T--){ Sca2(N,M); pre.clear(); pre[N] = 0; for(int i = 1; i <= M ; i ++){ int l = read() + 1,r = read() + 1; q[i] = mp(l,r); pre[l]++; pre[r + 1]--; } map<int,int>::iterator it; int ans = 0; int sum = 0; int la = 0; for(it = pre.begin(); it != pre.end(); it++){ PII t = *it; if(sum) ans += t.fi - la - 1; la = t.fi; sum = (sum + t.se) & 1; if(sum) ans++; } printf("Case #%d: ",Case++); Pri(ans); } return 0; }
C.unsolved by hl
小范围n²logn的树状数组常规暴力
大范围三次FFT解决a,b,c三个元作为最大值的情况，加上a,b,c三个元有两个同时作为最大值的情况，加上三个数一样的情况即可.
#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 = 4e5 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; LL a[maxn],b[maxn],c[maxn]; struct complex{ double x,y; complex(){} complex(double x,double y):x(x),y(y){} inline friend complex operator - (complex a,complex b){return complex(a.x - b.x,a.y - b.y);} inline friend complex operator + (complex a,complex b){return complex(a.x + b.x,a.y + b.y);} inline friend complex operator * (complex a,complex b){return complex(a.x * b.x - a.y * b.y,a.x * b.y + a.y * b.x);} }A[maxn << 2],B[maxn << 2]; int r[maxn << 2],limit; inline void FFT(int limit,complex *A,int *r,int type){ for(register int i = 0 ; i < limit; i ++) if(i < r[i]) swap(A[i],A[r[i]]); for(register int mid = 1; mid < limit; mid <<= 1){ complex Wn(cos(PI / mid),type * sin(PI / mid)); int len = mid << 1; for(register int j = 0; j < limit; j += len){ complex w(1,0); for(register int k = 0 ; k < mid; k ++, w = w * Wn){ complex x = A[j + k],y = w * A[j + mid + k]; A[j + k] = x + y; A[j + mid + k] = x - y; } } } } LL pre[maxn]; LL tmp[maxn],cnt[maxn]; LL x[maxn],y[maxn],z[maxn]; inline LL cul(LL *a,LL *b,LL *c){ for(register int i = 0 ; i <= M ; i ++) tmp[i] = cnt[i] = 0; for(register int i = 1; i <= N ; i ++) tmp[b[i]]++; for(register int i = 1 ; i <= M ; i ++) tmp[i] += tmp[i - 1]; for(register int i = 1; i <= N ; i ++) cnt[a[i]]++; LL sum = 0; for(register int i = 1; i <= N ; i ++) sum += cnt[c[i]] * tmp[c[i] - 1]; return sum; } inline LL solve(LL *a,LL *b,LL *c){ LL ans = 0; int cnt1 = 1,cnt2 = 1; for(register int i = 0 ; i < limit; i ++) A[i].y = B[i].y = A[i].x = B[i].x = 0; for(register int i = 1; i <= N ; i ++) A[b[i]].x++; for(register int i = 1; i <= N ; i ++) B[c[i]].x++; FFT(limit,A,r,1); FFT(limit,B,r,1); for(register int i = 0; i < limit ; i ++) A[i] = A[i] * B[i]; FFT(limit,A,r,-1); pre[limit] = 0; for(register int i = 0 ; i < limit ; i ++) pre[i] = (LL)(A[i].x / limit + 0.5); for(register int i = limit - 1; i >= 0; i --) pre[i] += pre[i + 1]; for(register int i = 1; i <= N ; i ++){ while(cnt1 <= N && b[cnt1] < a[i]) cnt1++; while(cnt2 <= N && c[cnt2] < a[i]) cnt2++; LL v = N + 1 - cnt1,u = N + 1 - cnt2; ans += pre[a[i]] - v * N - u * N + v * u; } return ans; } LL tree[maxn]; inline void add(int u,int x){ for(;u <= M + M; u += u & -u) tree[u] += x; } inline int getsum(int x){ int ans = 0; for(;x > 0; x -= x & -x) ans += tree[x]; return ans; } inline LL solve2(LL *a,LL *b,LL *c){ for(int i = 1; i <= N ; i ++){ add(c[i],1); } LL ans = 0; for(int i = 1; i <= N ; i ++){ for(int j = 1; j <= N ; j ++){ int r = a[i] + b[j],l = abs(a[i] - b[j]); ans += getsum(r); if(l > 0) ans -= getsum(l - 1); } } for(int i = 1; i <= N ; i ++) add(c[i],-1); return ans; } int main(){ int T = read(); int Case = 1; while(T--){ Sca(N); M = 0; for(int i = 1; i <= N ; i ++) a[i] = read(),M = max((LL)M,a[i]); for(int i = 1; i <= N ; i ++) b[i] = read(),M = max((LL)M,b[i]); for(int i = 1; i <= N ; i ++) c[i] = read(),M = max((LL)M,c[i]); sort(a + 1,a + 1 + N); sort(b + 1,b + 1 + N); sort(c + 1,c + 1 + N); if(N <= 1000){ printf("Case #%d: ",Case++); LL ans = solve2(a,b,c); Prl(ans); continue; } limit = 1; int l = 0; while(limit <= M + M) limit <<= 1,l++; for(int i = 0 ; i < limit; i ++) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1)); LL ans = 0; ans += solve(a,b,c); ans += solve(b,c,a); ans += solve(c,a,b); LL sum = 0; sum += cul(a,b,c); sum += cul(b,c,a); sum += cul(c,a,b); ans += sum; for(int i = 0 ; i <= M ; i ++) x[i] = y[i] = z[i] = 0; for(int i = 1; i <= N ; i ++){ x[a[i]]++; y[b[i]]++; z[c[i]]++; } for(int i = 0; i <= M ; i ++) ans += x[i] * y[i] * z[i]; printf("Case #%d: ",Case++); Prl(ans); } return 0; }
D.12:43:08 solved by zcz
#include<iostream> #include<cstring> #include<cstdio> #include<vector> using namespace std; long long ans[3005]; int vis[3005]; long long A[3005]; long long maxn=3000; long long mod=1e9+7; vector<long long> v; long long q_p(long long a,long long b){ long long ans = 1; while(b){ if(b & 1)ans = ans * a % mod; b >>= 1; a = a * a % mod; } return ans; } long long inv(long long a){ return q_p(a,mod - 2); } void dfs(long long sum,long long s,long long t) { long long cnt1=s-sum; if(cnt1+v.size()+1>maxn) return ; v.push_back(t); vis[t]++; long long fz=0; long long fm=1; for(int i=0;i<v.size();i++) { if(i==0||v[i]!=v[i-1]) { fm*=A[vis[v[i]]]; fm%=mod; } } fm*=A[cnt1]; fm%=mod; fz=A[cnt1+v.size()]; ans[cnt1+v.size()]+=fz*inv(fm); ans[cnt1+v.size()]%=mod; for(int i=t;i<=maxn;i++) { dfs(sum+i,s*i,i); } v.pop_back(); vis[t]--; } int main() { A[0]=1; for(int i=1;i<=maxn;i++) { A[i]=A[i-1]*i; A[i]%=mod; } for(int i=2;i<=maxn;i++) { dfs(i,i,i); } int T; cin>>T; while(T--) { int n; cin>>n; cout<<ans[n]<<endl; } return 0; }
F.3:38:14(-2) solved by gbs
#include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> #include <map> #include <set> #include <iostream> #include <string> #include <algorithm> #include <vector> #include <queue> using namespace std; typedef __int128 LL; void scan(__int128 &x)//输入 { x = 0; int f = 1; char ch; if((ch = getchar()) == '-') f = -f; else x = x*10 + ch-'0'; while((ch = getchar()) >= '0' && ch <= '9') x = x*10 + ch-'0'; x *= f; } void print(__int128 x) { if (!x) return ; if (x < 0) putchar('-'),x = -x; print(x / 10); putchar(x % 10 + '0'); } char stra[30]; //LL dp[30][30][30]; //LL dp[30][30]; LL fdp[30][30][30]; LL sdp[30][30]; int main() { for (int j=1; j<=29; j++) sdp[0][j] = 1; for (int i=1; i<=28; i++) { if (i == 1) { //第一个数的最大值是j for (int j=1; j<=28; j++) fdp[i][j][j] = 1,fdp[i][j][j-1] = j-1,sdp[i][j] += j; continue; } for (int j=1; j+i-1<=28; j++)//防止爆26 { //cout<<"!!"<<i<<' '<<j<<endl; for (int k =max(1,j-1); k<=j+i-1; k++) { fdp[i][j][k] = fdp[i-1][j][k] * k + fdp[i-1][j][k-1]; //cout<<"@"<<fdp[i][j][k]<<endl; //if (i==2 && j==2) // cout<<k<<' '<<fdp[i][j][k]<<' '<<fdp[i-1][j][k]<<' '<<fdp[i-1][j][k-1]<<endl; sdp[i][j] += fdp[i][j][k]; } } } //cout<<sdp[24][1]<<endl;; //cout<<sdp[25][1]<<endl;; //print(sdp[26][1]); int T; int n; LL m; int casen = 1; cin >>T; while(T--) { cin>>n; getchar(); scan(m); stra[n] = 0; stra[0] = 'A'; //m--; int now_bit = 1;//目前位 int now_c; for (int i=1; i<n; i++) { now_c = 1; //cout<<"!!!!"<<n-1-i<<' '<<now_bit+1<<' '<<sdp[n-i-1][now_bit+1]<<endl; while(m > sdp[n-i-1][now_bit+1]) { //cout<<"("<<i<<' '<<now_bit<<' '<<sdp[n-i-1][now_bit]<<endl; m-=sdp[n-i-1][now_bit+1]; now_c++; now_bit = max(now_c,now_bit); //!!!!!!! } stra[i] = 'A' +now_c-1; //cout<<"!!"<<now_c<<stra[i]<<endl; } printf("Case #%d: %s\n",casen++,stra); } #ifdef VSCode system("pause"); #endif return 0; } /*for (int i=1; i<=26; i++) { dp[i][0][0] = 1; for (int j=1; j<= i; j++) { for (int k = 1;k<=j; k++) { dp[i][j][k] += dp[i][j-1][k] *k + dp[i][j-1][k-1]; //cout<<i<<' '<<j<<' '<<k<<" "<<dp[i][j][k]<<endl; } } } int n1; while(cin >>n1) { LL lltest = 0; for (int i=1; i<=n1; i++) lltest += dp[n1][n1][i],cout<<i<<"*"<<dp[n1][n1][i]<<endl;; cout<<lltest<<endl; }*/ /*dp[1][1] = 1; for (int i=2; i<=26; i++) { dp[i][1] = 1; for (int j=2; j<=i-1; j++) { dp[i][j] = dp[i-1][j-1] + dp[i-1][j]*j; } dp[i][i] = 1; } for (int i=1; i<=26; i++) { cout<<i<<":"<<dp[26][i]<<endl; }*/
I.3:33:45(-4) solved by zcz
#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; void scan(__int128 &x)//输入 { x = 0; int f = 1; char ch; if((ch = getchar()) == '-') f = -f; else x = x*10 + ch-'0'; while((ch = getchar()) >= '0' && ch <= '9') x = x*10 + ch-'0'; x *= f; } void _print(__int128 x) { if(x > 9) _print(x/10); putchar(x%10 + '0'); } void print(__int128 x)//输出 { if(x < 0) { x = -x; putchar('-'); } _print(x); } int main() { int T; cin>>T; while(T--) { __int128 n,a,b; LL N,A,B; cin >> N >> A >> B; if(N==1) { puts("0"); continue; } n = N; a = A; b = B; __int128 ans=a+(n-1)*b; for(int i=2;i<=100;i++) { __int128 tem=1; __int128 di=max(pow(n,1.0/i)-2,2.0); for(int j=1;j<=i;j++) tem*=di; __int128 cnt=i*(di-1); for(int j=di+1;tem<n;j++) { for(int k=1;k<=i&&tem<n;k++) { tem/=(j-1); tem*=j; cnt++; } } ans=min(ans,a*i+cnt*b); if(cnt==i) break; } print(ans); puts(""); } return 0; }
J.00:24:30 solved by gbs
#include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> #include <map> #include <set> #include <iostream> #include <string> #include <algorithm> #include <vector> #include <queue> using namespace std; typedef long long LL; int an[100]; LL nowpower[100]; LL num[100]; int nowbit = 0; int T; int n,b; LL the_ans() { nowpower[0] = 0; num[0] = 1; LL ans = 0; LL before = 0; for (int i=1; i<=nowbit; i++) { nowpower[i] = nowpower[i-1] * b + (b-1)*b/2*num[i-1]; num[i] = num[i-1] *b; } for (int i = nowbit -1; i>=0; i--) { //cout<<":"<<i<<' '<<an[i]<<endl; int t1 = an[i]; if (t1 >0) { ans += t1 * nowpower[i]; ans += ((t1 -1)*t1/2 ) *num[i] + before * t1 * num[i]; before += t1; ans += t1; } } return ans; } int main() { cin >>T; int casen = 1; while(T--) { scanf("%d%d",&n,&b); int n1= n; nowbit = 0; while(n1) { //cout<<n1<<' '<<nowbit<<endl; an[nowbit++] = n1%b; n1/=b; } printf("Case #%d: %lld\n",casen++,the_ans()); } #ifdef VSCode system("pause"); #endif return 0; }
