/**[线段树]hdu 2430 Beans给出一系列数，求最大的连续子段和且满足sigma % p <= k10^6，只能想Nlog(N)的算法。分别枚举以i结尾的子段的最优值,则使得sum[i] - sum[id]最大即是使id最小。问题转化为了求最小id。易知,满足条件的sum[id] % p 的范围 为 [sum[i] % p - k,sum[i] % p],所以对取模后的结果建树,维护里面最小的id值。*/#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;#define N 1000002#define INF 10000000#define L(i) (i) << 1#define R(i) (i) << 1 | 1struct _st{ int l,r,Min; int mid() { return (l + r) >> 1; }}st[N<<2];void build(int id,int l,int r){ st[id].l = l; st[id].r = r; st[id].Min = INF; if(l == r) return; int mid = st[id].mid(); build(L(id),l,mid); build(R(id),mid+1,r);}int query(int id,int l,int r){ if(st[id].l >= l && st[id].r <= r) return st[id].Min; int mid = st[id].mid(); if(mid >= r) return query(L(id),l,r); else if(mid < l) return query(R(id),l,r); else return min(query(L(id),l,mid),query(R(id),mid+1,r));}void update(int id,int p,int val){ if(st[id].l == st[id].r) { if(val < st[id].Min) st[id].Min = val; return ; } int mid = st[id].mid(); if(mid >= p) update(L(id),p,val); else update(R(id),p,val); st[id].Min = min(st[L(id)].Min, st[R(id)].Min);}__int64 s[N];int main(){ int cas = 0; int t,n,p,k,i,w; scanf("%d",&t); while(t--) { scanf("%d%d%d",&n,&p,&k); build(1,0,p); for(i = 1; i <= n; ++i) { scanf("%d",&w); s[i] = s[i-1] + w; } __int64 res = -1; int id; for(i = 1; i <= n; ++i) { if(s[i] % p <= k) id = 0; else id = query(1,s[i] % p - k ,s[i] % p); if(id != INF) res = max(res,(s[i] - s[id])/p); update(1,s[i] % p,i); } printf("Case %d: %I64d\n",++cas,res); } return 0;}
