3.17模拟总结
今天终于弄明白了lemon评测的文件格式!
T1
经典的一道题,贪心:对于第一个元素,只能由第二个元素给他,给完之后,第二个元素就变成了第一个元素,一直循环下去就行
#include<cstdio> #include<queue> #include<iostream> #include<cstring> #include<algorithm> #include<cmath> #include<cctype> #include<vector> #include<string> using namespace 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
贪心,每次找到第一个单调减的数字,删掉
#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
要求最长不上升子序列个数,就是求最长上升序列的长度,树状数组维护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
这个就是线段覆盖问题,贪心的板子,注意这里的区间左闭右开,我就当成闭区间错了(
按右端点排序,如果当前左端点不小于之前最右端点,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
大体上和上一道题一样,右端点排序,但这回是闭区间,而且条件变成当前区间左端点不超过之前最右端点
#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); }