中国(北方)大学生程序设计训练赛(第一周)

Problem D: 数学题

题目链接:DDDDDD
取个倒数就可以变成乘法了… 二分答案 然后chick的时候 双指针
卡精度…要用 long double

#include <bits/stdc++.h>
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef vector<int> vi;
typedef long long ll;
typedef pair<int,int> pii;
const ll mod=1000000007;
const ll inf=(1LL<<60);
const double pi=acos(-1);
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
inline void pcas(int ca) {printf("Case %d: ",ca);}
const int maxn=1e5+10;
typedef long double ld;
ld a[maxn],b[maxn];
double bb[maxn],aa[maxn];
const ld esp=1e-8;
ll n,m,k;
bool ok(ld x)
{
    ll cnt=0,now=m;
    rep(i,1,n) {
        while(now) {
            if(a[i]*b[now]>x) now--;
            else break;
        }
        cnt+=now;
    }
    return (n*m-cnt)<k;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--) {
        scanf("%lld%lld%lld",&n,&m,&k);
        for(int i = 1; i <= n; ++i){
            scanf("%lf",&aa[i]);
            a[i]=aa[i];
        }
        for(int i = 1; i <= m; ++i) {
            scanf("%lf",&bb[i]);
            b[i]=(ld)1.0/bb[i];
        }
        sort(a+1,a+n+1);
        sort(b+1,b+m+1);
        ld l=a[1]*b[1],r=a[n]*b[m];
        while(r-l>esp) {
            ld mid=(l+r)/2.0;
            if(ok(mid)) r=mid;
            else l=mid;
        }
        printf("%.2Lf\n",l);
    }
    return 0;
}

Problem E: Water Problem

题目链接

矩阵快速幂 维护 f(x) f(x-1) sin(π*x/2) cos(π*x/2)
转移矩阵
1,1,0,0
1,0,0,0
0,0,0,-1
1,0,1,0

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;

typedef long long LL;
const int MAXN = 4;
const LL MOD = 1e9+7;
typedef long long LL;
typedef struct
{
    LL mat[MAXN][MAXN];
    void Init()
    {
        memset(mat, 0, sizeof(mat));
        for(int i=0; i<MAXN; i++)
            mat[i][i] = 1;
    }
} Matrix;
Matrix p = {1,1,0,0,
            1,0,0,0,
            0,0,0,-1,
            1,0,1,0
           };
Matrix Mul_Matrix(Matrix a, Matrix b)
{
    Matrix c;
    for(int i=0; i<MAXN; i++)
    {
        for(int j=0; j<MAXN; j++)
        {
            c.mat[i][j] = 0;
            for(int k=0; k<MAXN; k++)
            {
                c.mat[i][j] += (a.mat[i][k] * b.mat[k][j]+MOD) % MOD;
                c.mat[i][j] %= MOD;
            }
        }
    }
    return c;
}
Matrix quick_Mod_Matrix(LL m)
{
    Matrix ans, b = p;
    ans.Init();
    while(m)
    {
        if(m & 1)
            ans = Mul_Matrix(ans, b);
        m>>=1;
        b = Mul_Matrix(b, b);
    }
    return ans;
}
int main()
{
    int T;
    LL n, a, b;
   while(~scanf("%lld%lld%lld",&a,&b,&n)){
        if(n == 1)
        {
            printf("%lld\n",a);
            continue;
        }
        if(n == 2)
        {
            printf("%lld\n",b);
            continue;
        }
        Matrix tmp = quick_Mod_Matrix(n-2);
        LL ans = b*tmp.mat[0][0] % MOD;
        ans = (ans + a*tmp.mat[1][0]%MOD+MOD) % MOD;
        ans = (ans + 1*tmp.mat[2][0]%MOD+MOD) % MOD;
        printf("%lld\n",ans);
   }
return 0;
}

Problem F: 等差区间

—-题目链接—-
F. 区间 [L,R] 内的数排序后构成等差数列可分两种情况
1.公差为 0
2.公差不为 0 ⟺ 区间内无相同元素 且 相邻两项差构成的数列的GCD ×(R−L) = (区间最大值-区间最小值)
所以RMQ查询区间最大值最小值以及(各个数的上一个相同数的下标的最大值)以及区间GCD

#include <bits/stdc++.h>
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef vector<int> vi;
typedef long long ll;
typedef pair<int,int> pii;
const ll mod=1000000007;
const ll inf=(1LL<<60);
const double pi=acos(-1);
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
inline void pcas(int ca) {printf("Case %d: ",ca);}
const int maxn=1e5+10;
int n,q;
int a[maxn],dmx[maxn][30],dmi[maxn][25],pre[maxn][25],gcd[maxn][25];
int b[maxn],mat[maxn*10];
void RMQ_init()
{
    memset(mat,0,sizeof mat);
    for(int i = 0; i < n; ++i) dmx[i][0]=dmi[i][0]=a[i];
    for(int i = 0; i < n; ++i) {
        pre[i][0]=mat[a[i]];
        mat[a[i]]=i;
    }
    for(int j = 1; (1<<j)<=n; ++j)
    for(int i = 0; i +(1<<j)-1<n; ++i) {
        dmi[i][j]=min(dmi[i][j-1],dmi[i+(1<<(j-1))][j-1]);
        dmx[i][j]=max(dmx[i][j-1],dmx[i+(1<<(j-1))][j-1]);
        pre[i][j]=max(pre[i][j-1],pre[i+(1<<(j-1))][j-1]);
        gcd[i][j]=__gcd(gcd[i][j-1],gcd[i+(1<<(j-1))][j-1]);
    }
}
void RMQ(int l,int r,int &mx,int &mi,int &ok,int& gc)
{
    int k=0;
    while((1<<(k+1))<=r-l+1)k++;
    mx=max(dmx[l][k],dmx[r-(1<<k)+1][k]);
    mi=min(dmi[l][k],dmi[r-(1<<k)+1][k]);
    ok=max(pre[l][k],pre[r-(1<<k)+1][k]);
    l++;
    k=0;
    while((1<<(k+1))<=r-l+1)k++;
    gc=__gcd(gcd[l][k],gcd[r-(1<<k)+1][k]);
}
int main()
{

    while(~scanf("%d%d",&n,&q)){
        rep(i,0,n-1) {
            scanf("%d",&a[i]);
            if(i) gcd[i][0]=abs(a[i]-a[i-1]);
        }
        RMQ_init();
        int l,r,mx,mi,sub1,sub2;
        while(q--) {
            scanf("%d%d",&l,&r);
            l--,r--;
            int ok,temp;
            RMQ(l,r,mx,mi,ok,temp);
            if(mx==mi||l==r) {
                puts("Yes");
            }
            else {
                if(ok<=l){
                    if((ll)temp*(r-l)==(ll)(mx-mi)) puts("Yes");
                    else puts("No");
                }
                else puts("No");
            }
        }
    }
    return 0;
}
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