题目大意：有1~n型号的滑冰鞋，，每种有k双，一个x号脚的人只能适应[x,x+d]号滑冰鞋，每次增加一些x号脚的人或减少一些x号脚的人，问能否匹配
OTZ
这题我居然还能贡献一个WA真是醉了
#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define M 200200using namespace std;struct abcd{long long l_max,r_max,_max,sum;abcd() {}abcd(long long _){l_max=max(_,0ll);r_max=max(_,0ll);_max=max(_,0ll);sum=_;}friend abcd operator + (const abcd &x,const abcd &y){abcd re;re.l_max=max(x.l_max,x.sum+y.l_max);re.r_max=max(y.r_max,y.sum+x.r_max);re._max=max(max(x._max,y._max),x.r_max+y.l_max);re.sum=x.sum+y.sum;return re;}};long long n,m,k,d;struct Segtree{Segtree *ls,*rs;abcd status;void* operator new (size_t){static Segtree mempool[M<<1],*C=mempool;return C++;}void Build_Tree(int x,int y){int mid=x+y>>1;if(x==y){new (&status)abcd(-k);return ;}(ls=new Segtree)->Build_Tree(x,mid);(rs=new Segtree)->Build_Tree(mid+1,y);status=ls->status+rs->status;}void Modify(int x,int y,int pos,int val){int mid=x+y>>1;if(x==y){new (&status)abcd(status.sum+val);return ;}if(pos<=mid)ls->Modify(x,mid,pos,val);elsers->Modify(mid+1,y,pos,val);status=ls->status+rs->status;}}*tree=new Segtree;int main(){int i,x,y;cin>>n>>m>>k>>d;tree->Build_Tree(1,n);for(i=1;i<=m;i++){scanf("%d%d",&x,&y);tree->Modify(1,n,x,y);puts(tree->status._max<=k*d?"TAK":"NIE");}return 0;}
当你开展的事业从事的行动穷途末路大势已去的时候，
