2015 Multi-University Training Contest 6
hdu 5363
/******************hdu 5363 **********************
/*求一个集合元素里偶数和的子集个数
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int N = 1000005 + 10;
const int MOD=1000000007;
LL t,n,m;
LL poww(LL a,LL b,LL MOD)
{
LL res=1;
while(b)
{
if(b&1) res=(res*a)%MOD;
a=a*a%MOD;
b>>=1;
}
return res;
}
int main()
{
while(~scanf("%I64d",&t))
{
for(int i=0; i<t; ++i)
{
scanf("%I64d",&n);
m=poww(2,n-1,MOD);
m+=MOD;
printf("%I64d\n",(m-1)%MOD);
}
} return 0;
}
hdu 5360
【解题思路】：
结构体 + 优先队列，首先第一个人的L[i]必须是0，然后在同样的下标为L[i]的进队列，然后队列中的元素按照r[i]的大小排序，使得R[i]最小的优先
/*题意：现在有n个人，然后soda想要n个人尽可能多的去野营，每个人去野营是要在soda询问他时，
每个人所知道的去的人数大于等于L[i] 小于等于R[i] 然后问哪种询问顺序，可以使去的人数最多
*/
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <bits/stdc++.h>
using namespace std;
const int N=1000005;
struct item{
int ll,rr,id;
bool operator < (const item &aa) const{
return rr>aa.rr;
}
item(){
ll=0;
rr=0;
id=0;
}
}s[N];
bool cmp(item a,item b){
if(a.ll!=b.ll) return a.ll<b.ll;
return a.rr<b.rr;
}
int num[N];
int main(){
//freopen("1.txt","r",stdin);
int T,n;scanf("%d",&T);
while(T--){
scanf("%d",&n);
for(int i=0; i<n; ++i) scanf("%d",&s[i].ll);
for(int i=0; i<n; ++i) scanf("%d",&s[i].rr),s[i].id=i+1;
sort(s,s+n,cmp);
/* cout<<"bug begin!!!!!!!!!!!!!!!"<<endl;
for(int i=0; i<n; ++i)
cout<<"s[i].id="<<s[i].id<<" "<<"s[i].ll="<<s[i].ll<<" "<<"s[i].rr="<<s[i].rr<<endl;
cout<<"bug end!!!!!!!!!!!!!!!"<<endl;
*/
priority_queue <item>val;
while(!val.empty()) val.pop();
if(s[0].ll!=0){ ///第一个元素不符合
puts("0");
for(int i=0; i<n; ++i)
printf("%d%c",i+1,i==n-1?'\n':' ');
continue;
}
int i=1,cur=0,c1=0,c2=n-1;///cur：邀请人个数，c1:接受人数
val.push(s[0]);
while(!val.empty()){
// cout<<"bug begin!!!!!!!!!!!!!!!"<<endl;
// cout<<"cur="<<cur<<" pre c1="<<c1<<" "<<"val.top().id="<<val.top().id<<" "<<"val.top().ll="<<val.top().ll<<"val.top().rr="<<val.top().rr<<endl
// cout<<"bug end!!!!!!!!!!!!!!!"<<endl;
for(;s[i].ll==cur&&i<n;++i) val.push(s[i]);
while(!val.empty()){
if(val.top().rr>=cur){
int t=val.top().id;
num[c1++]=t;
cur++;
val.pop();
break;
}
else num[c2--]=val.top().id;
val.pop();
}
for(;s[i].ll==cur&&i<n; ++i) val.push(s[i]);
// cout<<"last c1="<<c1<<endl;
}
// cout<<"zhong ce c1="<<c1<<endl;
for(; i<n; ++i) num[c2--]=s[i].id;
printf("%d\n",c1);
for(int i=0; i<n; ++i) printf("%d%c",num[i],i==n-1?'\n':' ');
} return 0;
}
/****************hdu 5355 ******************
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int N =2e5 + 10;
const int MOD=1000000007;
bool vis[N];
int ans[N];
int t,n,m;
int main()
{
scanf("%d",&t);
while(t--){
scanf("%d %d",&n,&m);
memset(ans,0,sizeof(ans));
int sum=((n+1)*n)/2;
int mm=sum/m;
if(sum%m==0&&mm>m){
puts("YES");
memset(vis,false,sizeof(vis));
while(m--){
int su=0;
int cnt=0;
for(int i=n;i>=1;i--){
if(!vis[i]&&su+i<=mm){
vis[i]=true;
ans[cnt++]=i;
su+=i;
if(su==mm){
break;
}
}
}
sort(ans,ans+cnt);
printf("%d %d",cnt,ans[0]);
for(int i=1; i<cnt; ++i) printf(" %d",ans[i]);
puts("");
}
}
else puts("NO");
} return 0;
}