2019 ICPC 上海

时间:2019-09-15
本文章向大家介绍2019 ICPC 上海，主要包括2019 ICPC 上海使用实例、应用技巧、基本知识点总结和需要注意事项，具有一定的参考价值，需要的朋友可以参考一下。

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;
}
B

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;
}
C

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;
}
D

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;
    }*/
F

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;
}
I

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;
}
J

原文地址：https://www.cnblogs.com/Hugh-Locke/p/11523676.html

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