BestCoder Round #18

1001 Primes Problem

#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\n",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\n",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\n",t1);
}
else{
printf("%.2lf\n",max(t1,t2));
}
}
continue;
}
if(!gao()){
printf("%.2lf\n",max(t1,t2));
continue;
}
else{
double t3 = qiu(x1) ;double t4 = qiu(x2);
t3 = max(t3,t4);
t1 = max(t1,t2);
printf("%.2lf\n",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\n",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\n",val[i]);
}
}
return 0;
}