清北考前刷题day2早安

/*
做法一：按h sort一遍，对于一段区间[i,j]，高度花费就是h[j]-h[i]
然后枚举区间，把区间内C排序，一个一个尽量选即可。
n^3logn
标算：n^3 dp
高度排序，保证从前往后调。
f[i][j]表示当前在第i栋楼，已经跳了j次楼的最小话费。转移枚举下一次跳那座楼。
f[k][j+1]=min(f[i][j],f[i][j]+h[k]-h[i]+c[k]);最后枚举f[i][j]<T，ans=max(ans,j)。
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

#define N 510

using namespace std;
int T,n,ans,cnt;
int f[N][N];//当前处于第i座楼房已经跳了j次楼最小花费 
struct node{
    int h,c;
    bool operator < (const node a) const{
            return h<a.h;
    }
    
}L[N];

inline int read()
{ 
    int x=0,f=1;char c=getchar();
    while(c>‘9‘||c<‘0‘){if(c==‘-‘)f=-1;c=getchar();}
    while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();}
    return x*f;
}

int main()
{
    n=read();
    for(int i=1;i<=n;i++) L[i].c=read();
    for(int i=1;i<=n;i++) L[i].h=read();
    T=read();
    sort(L+1,L+n+1);
    
    memset(f,127/3,sizeof f);
    f[1][0]=f[1][1]=L[1].c;
    for(int i=1;i<=n;i++)
      for(int j=0;j<=n+1;j++)
        for(int k=i+1;k<=n;k++)
          f[k][j+1]=min(f[k][j+1],f[i][j]+L[k].h-L[i].h+L[k].c);
                
    for(int i=1;i<=n;i++)
      for(int j=0;j<=n+1;j++)
        if(f[i][j]<=T) ans=max(ans,j);
    printf("%d\n",ans);
    return 0;
     
}

/*
考虑最后要求的数 a1,a2,..an假定从小到大
给定了b1,b2......bn 从小到大排序
性质一：a1+a2=b1
性质二：a1+a3=b2
假设a2+a3=x，可以解出a1,a2,a3,并可以在b中b1,b2,x删除。则剩下的最小的一定是a1+a4。由于a1已知，a4可解。这个过程可以反复下去。
枚举a2+a3=b里面哪个数就好
*/
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>

using namespace std;

const int N=310;

int n,m,res[N],ans[N][N],z[N*N],cnt;

bool use[N*N];

void check(int p)
{
    memset(use,false,sizeof(use));
    if ((z[1]+z[2]+z[p])&1) return;
    res[1]=(z[1]+z[2]+z[p])/2-z[p];
    res[2]=z[1]-res[1];
    res[3]=z[2]-res[1];
    use[1]=use[2]=use[p]=true;
    for (int a=4,b=3;a<=n;a++)
    {
        while (b<=m && use[b]) b++;
        if (b>m) return;
        res[a]=z[b]-res[1];
        use[b]=true;
        for (int c=2;c<a;c++)
        {
            if (res[c]>res[a]) return;
            int v=res[c]+res[a];
            int p=lower_bound(z+1,z+m+1,v)-z;
            if (z[p]!=v) return;
            int px=p;
            while (px && z[px]==z[p]) px--;
            px++;
            while (px<=m && z[px]==z[p] && use[px]) px++;
            if (z[px]!=z[p] || use[px]) return;
            p=px;    use[p]=true;
        }
    }
    cnt++;
    for (int a=1;a<=n;a++) ans[cnt][a]=res[a];
}

int main()
{
    freopen("city.in","r",stdin);
    freopen("city.out","w",stdout);
    scanf("%d",&n);
    m=n*(n-1)/2;
    for (int a=1;a<=m;a++) scanf("%d",&z[a]);
    sort(z+1,z+m+1);
    for (int a=3;a<=m;)
    {
        check(a);int b=a;
        while (b<=m && z[b]==z[a])b++;
        a=b;
    }
    printf("%d\n",cnt);
    for (int a=1;a<=cnt;a++)
        for (int b=1;b<=n;b++)
        {
            printf("%d",ans[a][b]);
            if (b==n) printf("\n");
            else printf(" ");
        }
    return 0;
}

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn = 100009;
const int maxv = 10000;
const int bsz = 100;
const int maxb = 103;

int n, m;
int a[maxn], vb[maxb][maxb], ve[maxb][maxb];
int xb[maxn], xe[maxn];
int i_buf[maxn * maxb * 2], tib;

void pre() {
    memset(ve, 0, sizeof(ve));
    memset(xe, 0, sizeof(xe));
    for (int i = 1; i <= n; ++ i)
        ++ xe[a[i]];
    for (int i = 0; i <= maxv; ++ i) {
        xb[i] = tib;
        tib += xe[i];
        xe[i] = xb[i];
    }
    for (int i = 1; i <= n; ++ i)
        i_buf[xe[a[i]] ++] = i;
    for (int m = 1; m <= bsz; ++ m) {
        for (int i = 1; i <= n; ++ i)
            ++ ve[m][a[i] % m];
        for (int i = 0; i < m; ++ i) {
            vb[m][i] = tib;
            tib += ve[m][i];
            ve[m][i] = vb[m][i];
        }
        for (int i = 1; i <= n; ++ i)
            i_buf[ve[m][a[i] % m] ++] = i;
    }
}

int queryb(int l0, int r0, int p, int k) {
    if (vb[p][k] == ve[p][k])
        return 0;
    int *x1 = lower_bound(i_buf + vb[p][k], i_buf + ve[p][k], l0);
    int *x2 = upper_bound(i_buf + vb[p][k], i_buf + ve[p][k], r0);
    return x2 - x1;
}

int querys(int v, int l0, int r0) {
    if (xb[v] == xe[v])
        return 0;
    int *x1 = lower_bound(i_buf + xb[v], i_buf + xe[v], l0);
    int *x2 = upper_bound(i_buf + xb[v], i_buf + xe[v], r0);
    return x2 - x1;
}

int querya(int l0, int r0, int p, int k) {
    int ans = 0;
    for (int i = k; i <= maxv; i += p)
        ans += querys(i, l0, r0);
    return ans;
}

int main() {
    freopen("light.in", "r", stdin);
    freopen("light.out", "w", stdout);

    scanf("%d%d", &n, &m);
    tib = 0;
    for (int i = 1; i <= n; ++ i)
        scanf("%d", a + i);
    pre();
    while (m --) {
        int l, r, p, k;
        scanf("%d%d%d%d", &l, &r, &p, &k);
        if (p <= bsz)
            printf("%d\n", queryb(l, r, p, k));
        else
            printf("%d\n", querya(l, r, p, k));
    }
}