2015多校赛第六场

1008：Hiking

题意：There are nn people
and ii-th
people will go out if the number of people going out if no less than l_{i}
+ 1l​i​​+1 and
no larger than r_{i}
+ 1r​i​​+1.
You need to invite them one by one to make the number of people going out as large as possible.

思路：贪心做。。对于当前我们可以邀请的人，先邀请 r 小的，然后维护当前可以邀请的人就行了，把当前能邀请的然放进一个优先队列，每次取出 r 最小的，但是 r 得不小于当前已经参加的人数。至于怎么找当前可以邀请的人， 先把所有的[l, r]按 l 升序排列，，然后逐个添加到优先队列里就好了。所以会有两次排序，，一次按
l ，一次按 r

#pragma warning(disable:4996)
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#define N 100010
using namespace std;

struct line{
int l, r, id;
bool operator<(const line&op)const{
return r > op.r;
}
};

bool cmp(const line&a, const line&b){
if (a.l == b.l)return a.r < b.r;
return a.l < b.l;
}

line a
;
bool vis
;
int ans
;
priority_queue<line>q;

int main(){
int t;
scanf("%d", &t);
while (t--){
int n;
scanf("%d", &n);
memset(vis, false, sizeof vis);
memset(ans, 0, sizeof ans);
while (!q.empty())q.pop();
for (int i = 1; i <= n; i++)scanf("%d", &a[i].l);
for (int i = 1; i <= n; i++){
scanf("%d", &a[i].r);
a[i].id = i;
}

sort(a + 1, a + 1 + n, cmp);

int cnt = 0;

int j = 1;
for (int i = 0; i < n; i++){
for (; j <= n; j++){
if (a[j].l <= i){
q.push(a[j]);
}
else break;
}
if (q.empty())break;
line now = q.top();
while (now.r < i){
q.pop();
if (q.empty()){
break;
}
now = q.top();
}
if (q.empty())break;
ans[i] = now.id;
vis[now.id] = true;
cnt++;
q.pop();
}
if (ans[0] == 0){
printf("0\n1");
for (int i = 2; i <= n; i++)printf(" %d", i);
printf("\n");
}
else{
printf("%d\n%d", cnt, ans[0]);
for (int i = 1; i < cnt; i++)printf(" %d", ans[i]);
for (int i = 1; i <= n; i++){
if (vis[i] == false)printf(" %d", i);
}
printf("\n");
}

}
return 0;
}

1011：Key Set

题意：For a given set {1, 2, . . . , nn},
you are to find the number of nonempty subsets with even sum.

思路：答案就是2^( n- 1) - 1。。至于为什么，找规律也行。规范点的方法就是考虑{2,3,4，......n}的集合。共有2^(
n- 1) - 1个非空的子集，然后

对于每一个子集，若sum为偶数，就不管了，若sum为奇数就把
1 加进去，就成偶数了。O(∩_∩)O

#include <iostream>
#include <cstdio>

using namespace std;

long long pow( long long n )
{
long long ans = 1;
long long base = 2;
while( n )
{
if( n&1 )
ans = base*ans%1000000007;
base = base * base %1000000007;
n>>=1;
}
return ans;
}

int main()
{
int t;
cin >> t;
while( t-- )
{
long long n;
scanf("%I64d", &n);
long long ans = pow( n-1 ) -1;
printf("%I64d\n", ans);
}
return 0;
}