BestCoder Round #18

1001 Primes Problem

 打个10^4的素数表，枚举前两个搞一下就好了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<vector>
#include<stdlib.h>
#include<algorithm>
#include<math.h>
using namespace std;
const int n = 10000;
int vis[11111];
int cnt ;
int ans[11111];
void gao()
{
    memset(vis,0,sizeof(vis));
    for(int i= 2;i<=sqrt(n);i++){
        for(int j= 2;i*j<=n;j++)
            if(!vis[i]) vis[i*j] = 1;
    }
    vis[0] = 1;vis[1] = 1;
    cnt = 0;
    for(int i = 2;i<=n;i++)
        if(!vis[i]) ans[cnt++] = i;
}

int main()
{
    gao();
    int k;
    while(cin>>k){
        int sum = 0;
        for(int i = 0;i<cnt;i++){
            if(ans[i]>=k) continue;
            for(int j=i;j<cnt;j++){
                int t = k-ans[i]-ans[j];
                if(t<=0) break;
                if(vis[t]||t<ans[j]) continue;
                else sum++;
            }
        }
        printf("%d
",sum);
    }
    return  0;
}

1002 Math Problem
解方程啊，预测数据太水，我没加根号都能过预测，无语。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<vector>
#include<stdlib.h>
#include<algorithm>
#include<math.h>
using namespace std;
#define eps  = 1e-7
double a,b,c,d,L,R;
double x1,x2;
int gao()
{
    double  t =  4*b*b - 12*a*c;
    if(t<0) return 0;
    else{
        x1 = (-2) * b + sqrt(t);
        x1/=6*a;
        x2 = (-2)*b - sqrt(t); x2/=6*a;
    }
    return 1;
}

double qiu(double x)
{
    double t = a*x*x*x + b*x*x + c*x +d;
    return fabs(t);
}
double qiu1(double x)
{
    double t = b*x*x + c*x+d;
    return fabs(t);
}
int main()
{
    double t1,t2;
    while(cin>>a>>b>>c>>d>>L>>R){
        t1 = qiu(L);t2 = qiu(R);
        if(a==0){
            if(b==0){
                printf("%.2lf
",max(t1,t2));
            }
            else{
                double t4 = (-c) / (2*b);
                if(t4>=L&&t4<=R){
                double t3 = qiu1((-c)/(2*b));
                t1 = max(t1,t2);
                t1 =max(t1,t3);
                printf("%.2lf
",t1);
                }
                else{
                    printf("%.2lf
",max(t1,t2));
                }
            }
            continue;
        }
        if(!gao()){
            printf("%.2lf
",max(t1,t2));
            continue;
        }
        else{
            double t3 = qiu(x1) ;double t4 = qiu(x2);
            t3 = max(t3,t4);
            t1 = max(t1,t2);
            printf("%.2lf
",max(t1,t3));
        }
    }
    return 0;
}

1003 Bits Problem
数位dp，记录一个的到当前位满足条件的1的种类数，记录一个到当前位满足条件的数的和
因为是开区间，最后单独判断一下右端点。

#include<stdio.h>
#include<string.h>
#include<vector>
#include<iostream>
#include<map>
#include<queue>

using namespace std;

const int mod = 1e9+7;
typedef long long LL;
const int maxn = 1111;
int dp_sum[maxn][maxn];
int dp_mod[maxn][maxn];
int n;
int val[maxn];
char str[maxn];
int dfs_sum(int x,int sum,int flag)
{
    if(~dp_sum[x][sum]&&!flag) return dp_sum[x][sum];
    int bound = flag? val[x]:1;
    if(x==0){
        if(sum==0)return 1;
        else return 0;
    }
    int ans = 0 ;
    for(int i = 0 ;i<=bound;i++){
        if(i==0) ans+=dfs_sum(x-1,sum,flag&&bound==i),ans%=mod;
        else{
            if(sum>0) ans+=dfs_sum(x-1,sum-1,flag&&bound==i),ans%=mod;
        }
    }
    return flag? ans:dp_sum[x][sum] = ans;
}

LL cifang(LL a,int b)
{
    LL ans = 1;
    while(b){
        if(b&1) ans*=a,ans%=mod;
        a*=a;a%=mod; b>>=1;
    }
    return ans;
}

LL dfs_mod(int x,int sum,int flag)
{
    if(~dp_mod[x][sum]&&!flag) return dp_mod[x][sum];
    int bound = flag? val[x] : 1;
    if(x==0) return 0;
    LL ans = 0 ;
    for(int i = 0 ;i<=bound;i++){
        if(i==0){
            ans+=dfs_mod(x-1,sum,flag&&bound==i),ans%=mod;
        }
        else{
            if(sum<=0) continue;
            int t = dfs_sum(x-1,sum-1,flag&&bound == i );
            ans+=t%mod*cifang(2,x-1)%mod;
            ans+=dfs_mod(x-1,sum-1,flag&&bound==i),ans%=mod;
        }
    }
    return flag?ans:dp_mod[x][sum] = ans;
}

void gao()
{
    int len =strlen(str);
    for(int i = 0 ;i<len;i++){
        val[i+1] = str[len-i-1] - '0';
    }
    LL t = dfs_mod(len,n,1);
    int ans = 0;
    int cnt = 0;
    for(int i = len;i>=1;i--)
        if(val[i]) cnt++;
    for(int i = len;i>=1;i--){
        ans=ans*2+val[i];ans%=mod;
    }
    if(cnt==n) t-=ans;
    if(t<0) t = (t+mod) % mod;
    printf("%I64d
",t);
}

int main()
{
    memset(dp_sum,-1,sizeof(dp_sum));
    memset(dp_mod,-1,sizeof(dp_mod));
   // system("pause");
    while(scanf("%d",&n)!=EOF){
        scanf("%s",str);
        gao();
    }
    return 0;
}

1004 K-short Problem
先离散化。
k最大为10嘛，每个节点记录从小到大记录十个数。
按x从小到大，y从小到大，h从小到大
以y轴建树
离线更新查询同时进行 ，线段树上的操作暴力搞就好。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <set>
#include <bitset>
#include <queue>
#include <stack>
#include <string>
#include <iostream>
#include <cmath>
#include <climits>

using namespace std;

#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
const int maxn = 66666;

vector<int> node[maxn<<2];
vector<int> gg;
int n,m;
vector<int> ans;
int val[maxn];
struct Node
{
    int x;int y;int h;int pos;
}edge[maxn],edge1[maxn];
int len;
int cmp(const Node &a,const Node &b)
{
    if(a.x!=b.x) return a.x<b.x;
    if(a.y!=b.y) return a.y<b.y;
    return a.h<b.h;
}

void gao1()
{
    gg.clear();
    for(int i = 0;i<n;i++)
        gg.push_back(edge[i].y);
    for(int i = 0;i<m;i++)
        gg.push_back(edge1[i].y);
    sort(gg.begin(),gg.end());
    gg.erase(unique(gg.begin(),gg.end()),gg.end());
    for(int i = 0;i<n;i++)
        edge[i].y = lower_bound(gg.begin(),gg.end(),edge[i].y) - gg.begin();
    for(int i =0;i<m;i++)
        edge1[i].y = lower_bound(gg.begin(),gg.end(),edge1[i].y) - gg.begin();
    len = gg.size() - 1;
}

void up(int rt)
{
    node[rt].clear();
    int a[22];int cnt = 0;
    for(int i= 0;i<node[rt<<1].size();i++){
        a[cnt++] = node[rt<<1][i];
    }
    for(int i =0;i<node[rt<<1|1].size();i++){
        a[cnt++] = node[rt<<1|1][i];
    }
    sort(a,a+cnt);
    cnt = min(cnt,10);
    for(int i =0;i<cnt;i++)
        node[rt].push_back(a[i]);
}

void build(int l,int r,int rt)
{
    node[rt].clear();
    if(l==r) return ;
    int mid=(l+r)>>1;
    build(lson);
    build(rson);
}

void update(int key,int add,int l,int r,int rt)
{
    if(l==r){
        node[rt].push_back(add);
        sort(node[rt].begin(),node[rt].end());
        return ;
    }
    int mid=(l+r)>>1;
    if(key<=mid) update(key,add,lson);
    else update(key,add,rson);
    up(rt);
}

void ask(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R){
        for(int i = 0 ;i<node[rt].size();i++)
            ans.push_back(node[rt][i]);
        return ;
    }
    int mid=(l+r)>>1;
    if(L<=mid) ask(L,R,lson);
    if(R>mid) ask(L,R,rson);
}

void gao(int L,int R,int k,int l,int r,int rt,int pos)
{
    ans.clear();
    ask(L,R,l,r,rt);
    if(ans.size()<k) val[pos] = -1;
    else{
        sort(ans.begin(),ans.end());
        val[pos] = ans[k-1];
    }
}
int main()
{
    while(cin>>n>>m){
        for(int i = 0 ;i<n;i++){
            scanf("%d%d%d",&edge[i].x,&edge[i].y,&edge[i].h);
        }
        for(int i = 0;i<m;i++){
            scanf("%d%d%d",&edge1[i].x,&edge1[i].y,&edge1[i].h);
            edge1[i].pos = i;
        }
        gao1();
        sort(edge,edge+n,cmp);
        sort(edge1,edge1+m,cmp);
        build(0,len,1);
        int cnt = 0;
        for(int i = 0;i<m;i++){
            while(1){
                if(cnt>=n) break;
                if(edge[cnt].x>edge1[i].x) break;
                update(edge[cnt].y,edge[cnt].h,0,len,1);
                cnt++;
            }
            gao(0,edge1[i].y,edge1[i].h,0,len,1,edge1[i].pos);
        }
        for(int i = 0;i<m;i++){
            printf("%d
",val[i]);
        }
    }
    return 0;
}