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;
}