BZOJ 1012 單調隊列+二分

思路：
維護一個單減的序列 序號是單增的 每回二分查找第一個比詢問的大的值
我手懶 用得lower_bound

//By SiriusRen
#include <cstdio>
#include <algorithm>
using namespace std;
#define int long long
int m,mod,top,jy,ans,tot;
char ch[3];
struct Node{int pos,weight;}node[200050],t;
bool operator < (Node a,Node b){return a.pos<b.pos;
}
signed main(){scanf("%lld%lld",&m,&mod);for(int i=1;i<=m;i++){scanf("%s%lld",ch,&jy);if(ch[0]=='A'){tot++,jy=(jy+ans)%mod;while(top&&node[top-1].weight<=jy)top--;node[top].pos=tot,node[top++].weight=jy;}else{t.pos=tot-jy+1;ans=lower_bound(node,node+top,t)->weight;printf("%lld\n",ans);}}
}