10.2 上午 考试
T1
移项，然后离散，dp即可
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <ctime>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define ll long long
using namespace std;
const int N=200006;
const int N2=500006;
struct LI
{
int pos,val,flag;
bool friend operator < (LI a,LI b)
{
return a.val<b.val;
}
}li[N2];
int con;
struct son
{
int x,w,v[2];
bool friend operator < (son a,son b)
{
return a.x<b.x;
}
}ji[N];
int n;
int f[N];
int ans;
int now;
void lisan()
{
for(int i=1;i<=n;++i)
{
li[++con]=(LI){i,ji[i].x-ji[i].w,0};
li[++con]=(LI){i,ji[i].x+ji[i].w,1};
}
sort(li+1,li+1+con);
li[0].val=-0x7fffffff;
for(int i=1;i<=con;++i)
{
if(li[i].val==li[i-1].val)
ji[li[i].pos].v[li[i].flag]=now;
else
ji[li[i].pos].v[li[i].flag]=++now;
}
}
void out11()
{
printf("\n");
for(int i=1;i<=n;++i)
printf("%d ",f[i]);
printf("\n");
}
int c[N2];
inline void add(int pos,int vv)
{
for(int i=pos;i<=now;i+=(i&(-i)) )
if(c[i]<vv)
c[i]=vv;
}
inline int qq(int pos)
{
int ans=0;
for(int i=pos;i>0;i-=(i&(-i)) )
if(ans<c[i])
ans=c[i];
return ans;
}
void dp()
{
int temp;
for(int i=1;i<=n;++i)
{
temp=qq(ji[i].v[0]);
f[i]=temp+1;
if(ans<f[i])
ans=f[i];
add(ji[i].v[1],f[i]);
}
//out11();
}
int main(){
//freopen("T1.in","r",stdin);
//freopen("T1.out","w",stdout);
scanf("%d",&n);
for(int i=1;i<=n;++i)
scanf("%d%d",&ji[i].x,&ji[i].w);
sort(ji+1,ji+1+n);
lisan();
dp();
cout<<ans;
}
T2
分块
#pragma GCC optimize("O2")
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <cmath>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define ll long long
using namespace std;
const int N=100006;
int n,m;
int a[N];
int fen,dui[N],L[1006],R[1006];
int mx[1006];
ll ji[1006];
void reset(int x)
{
mx[x]=ji[x]=0;
for(int i=L[x];i<=R[x];++i)
{
if(mx[x]<a[i])
mx[x]=a[i];
ji[x]+=a[i];
}
}
void mm(int l,int r,int mod)
{
int q1=min(R[dui[l]],r);
for(int i=l;i<=q1;++i)
a[i]%=mod;
reset(dui[l]);
if(dui[l]!=dui[r])
{
for(int i=L[dui[r]];i<=r;++i)
a[i]%=mod;
reset(dui[r]);
}
for(int i=dui[l]+1;i<dui[r];++i)
{
if(mx[i]<mod)
continue;
for(int j=L[i];j<=R[i];++j)
a[j]%=mod;
reset(i);
}
}
void change(int pos,int val)
{
a[pos]=val;
reset(dui[pos]);
}
ll qq(int l,int r)
{
ll ans=0;
int q1=min(R[dui[l]],r);
for(int i=l;i<=q1;++i)
ans+=a[i];
if(dui[l]!=dui[r])
for(int i=L[dui[r]];i<=r;++i)
ans+=a[i];
for(int i=dui[l]+1;i<dui[r];++i)
ans+=ji[i];
return ans;
}
int main(){
freopen("T2.in","r",stdin);
freopen("T2.out","w",stdout);
scanf("%d%d",&n,&m);
fen=(int)ceil(sqrt(n+0.5));
for(int i=1;i<=n;++i)
{
scanf("%d",&a[i]);
dui[i]=(i-1)/fen+1;
if(mx[dui[i]]<a[i])
mx[dui[i]]=a[i];
ji[dui[i]]+=a[i];
}
for(int i=1;i<=dui[n];++i)
{
L[i]=(i-1)*fen+1;
R[i]=min(n,i*fen);
}
int op,tin1,tin2,tin3;
for(int i=1;i<=m;++i)
{
scanf("%d",&op);
if(op==1)
{
scanf("%d%d",&tin1,&tin2);
if(tin1>tin2)
swap(tin1,tin2);
printf("%lld\n",qq(tin1,tin2));
}
else
if(op==2)
{
scanf("%d%d%d",&tin1,&tin2,&tin3);
if(tin1>tin2)
swap(tin1,tin2);
mm(tin1,tin2,tin3);
}
else
{
scanf("%d%d",&tin1,&tin2);
change(tin1,tin2);
}
}
}
T3
会炸long long INF 的SBT....
打表发现规律，然后 二分n即可
K=64，必须特判，不然炸了...
#include <cstdio>
#include <cstring>
#include <iostream>
#define mem(a,b) memset(a,b,sizeof(a))
#define ll long long
using namespace std;
ll num[123];
void chu(int x)
{
mem(num,0);
int temp=2*x;
for(int i=x+1;i<=temp;++i)
{
int cc=0;
for(int j=0;j<32;++j)
if( i&(1<<j) )
++cc;
++num[cc];
}
for(int i=1;i<=32;++i)
{
if(num[i]==0)
break;
printf("%lld ",num[i]);
}
printf(" ");
for(int i=7;i>=0;--i)
printf("%d",(((1<<i)&x)?1:0) );
printf("\n");
}
int main(){
freopen("biao.out","w",stdout);
for(int i=1;i<=100;++i)
chu(i);
}
#include <cstdio>
#include <cstring>
#include <iostream>
#define mem(a,b) memset(a,b,sizeof(a))
#define ll long long
using namespace std;
const ll INF=((ll)1<<63)-1;
int T,K;
ll m;
ll C[106][106];
void chu()
{
C[0][0]=1;
for(int i=1;i<66;++i)
C[i][0]=C[i][i]=1;
for(int i=1;i<66;++i)
for(int j=1;j<i;++j)
C[i][j]=C[i-1][j]+C[i-1][j-1];
}
ll ji(ll x)
{
ll ans=0;
int cc=0;
for(int i=63;i>=0;--i)
if( x&((ll)1<<i) )
{
//printf("i=%d\n",i);
++cc;
if(K-cc>=0)
ans+=C[i][K-cc];
}
return ans;
}
ll erl(ll l,ll r)
{
ll ans=INF,mid;
ll temp;
while(l<=r)
{
mid=(l+r)>>1;
temp=ji(mid);
if(temp==m&&ans>mid&&mid)
ans=mid;
if(l>=r)
break;
if(temp>=m)
r=mid-1;
else
l=mid+1;
}
return ans;
}
ll err(ll l,ll r)
{
ll ans=1,mid;
ll temp;
while(l<=r)
{
mid=(l+r)>>1;
//printf("l=%lld r=%lld mid=%lld\n",l,r,mid);
temp=ji(mid);
//printf("temp=%lld\n",temp);
if(temp==m&&ans<mid)
ans=mid;
if(l>=r)
break;
if(temp<=m)
l=mid+1;
else
r=mid-1;
}
return ans;
}
int main(){
//freopen("number10.in","r",stdin);
//freopen("in.in","r",stdin);
//freopen("out.out","w",stdout);
chu();
scanf("%d",&T);
ll l0,r0;
while(T--)
{
scanf("%lld%d",&m,&K);
if(K==0)
{
printf("0 1\n");
continue;
}
if(K==1)
{
printf("-1\n");
continue;
}
if(K==64)
{
printf("1 %lld\n",INF);
continue;
}
//K=5;
//printf("adsd=%lld\n",ji(97));
l0=erl(0,INF);
r0=err(0,INF);
printf("%lld %lld\n",l0,r0);
}
// cout<<(((ll)1<<63)-1);
}
