3.17模擬總結
阿新 • • 發佈:2022-03-17
今天終於弄明白了lemon評測的檔案格式!
T1
經典的一道題,貪心:對於第一個元素,只能由第二個元素給他,給完之後,第二個元素就變成了第一個元素,一直迴圈下去就行
#include<cstdio> #include<queue> #include<iostream> #include<cstring> #include<algorithm> #include<cmath> #include<cctype> #include<vector> #include<string> usingnamespace std; int n,a[105]; int sum,ans; int main(){ freopen("Playcard.in","r",stdin); freopen("Playcard.out","w",stdout); scanf("%d",&n); for(int i=1;i<=n;++i){ scanf("%d",a+i); sum += a[i]; } sum /= n; for(int i=1;i<=n;++i){ int x;if(a[i] != sum){ ans++; x = a[i] - sum; a[i] = sum; a[i+1] += x; } } printf("%d",ans); }
T2
貪心,每次找到第一個單調減的數字,刪掉
#include<cstdio> #include<queue> #include<iostream> #include<cstring> #include<algorithm> #include<cmath> #include<cctype> #include<vector> #include<string> using namespace std; int main(){ freopen("delete.in","r",stdin); freopen("delete.out","w",stdout); string a; int s; cin >> a >> s; int len = a.length(); for(int i=1;i<=s;++i){ for(int j=0;j<len;++j){ if(a[j] > a[j+1]){ for(int k=j;k<len-1;++k) a[k] = a[k+1]; break; } } } int flag = 1; for(int i=0;i<len-s;++i){ if(a[i] != '0')flag = 0; if(!flag)putchar(a[i]); } if(flag)putchar('0'); }
T3
要求最長不上升子序列個數,就是求最長上升序列的長度,樹狀陣列維護dp即可
#include<cstdio> #include<queue> #include<iostream> #include<cstring> #include<algorithm> #include<cmath> #include<cctype> #include<vector> #include<string> using namespace std; const int N = 30000 + 5; int c[N],a[1005],n,ans,mx; inline int lowbit(int x){ return x & (-x); } inline void update(int i,int x){ for(;i<=mx;i+=lowbit(i))c[i] = max(c[i],x); } inline int query(int i){ int res = 0; for(;i;i-=lowbit(i))res = max(res,c[i]); return res; } int main(){ freopen("missile.in","r",stdin); freopen("missile.out","w",stdout); int x; while(scanf("%d",&x) == 1){ a[++n] = x; mx = max(mx,x); } for(int i=1;i<=n;++i){ int now = query(a[i])+1; ans = max(ans,now); update(a[i]+1,now); } printf("%d",ans); }
T4
這個就是線段覆蓋問題,貪心的板子,注意這裡的區間左閉右開,我就當成閉區間錯了(
按右端點排序,如果當前左端點不小於之前最右端點,就能加入
#include<cstdio> #include<queue> #include<iostream> #include<cstring> #include<algorithm> #include<cmath> #include<cctype> #include<vector> #include<string> using namespace std; struct qwq{ int st,ed; bool operator <(qwq x)const{ if(ed != x.ed)return ed < x.ed; return st > x.st; } }a[1005]; int main(){ freopen("act.in","r",stdin); freopen("act.out","w",stdout); int n; scanf("%d",&n); for(int i=1;i<=n;++i)scanf("%d%d",&a[i].st,&a[i].ed); sort(a+1,a+1+n); int now_end = -1,ans = 0; for(int i=1;i<=n;++i){ if(a[i].st >= now_end){ now_end = a[i].ed; ans++; } } printf("%d",ans); }
T5
大體上和上一道題一樣,右端點排序,但這回是閉區間,而且條件變成當前區間左端點不超過之前最右端點
#include<cstdio> #include<queue> #include<iostream> #include<cstring> #include<algorithm> #include<cmath> #include<cctype> #include<vector> #include<string> using namespace std; struct qwq{ int x,y; bool operator <(qwq b)const{ return y < b.y; } }a[10005]; int main(){ freopen("a.in","r",stdin); freopen("a.out","w",stdout); int n; scanf("%d",&n); for(int i=1;i<=n;++i)scanf("%d%d",&a[i].x,&a[i].y); sort(a+1,a+1+n); int now = a[1].y; int ans = 1; for(int i=2;i<=n;++i){ if(now >= a[i].x)continue; ans++; now = a[i].y; } printf("%d",ans); }