这题目用线段树超时了，其实也差不多应该超时。10^6大数据量。看了一下网上的解法是单调队列。大概了解了一下，是个挺有意思的数据结构。
首先，需要求满足0<=(S[r]-S[l])%p<=k时，（S[r]-S[l]）的最大值。
由于S[r]>=S[l]，因此即求S[r]%p-k <= S[l]%p <= S[r]%p的最优解。
单调队列可解，按S[i]%p和i排列，每次固定r，求最优的l值，l在队头。
1 /* 2430 */ 2 #include <iostream> 3 #include <sstream> 4 #include <string> 5 #include <map> 6 #include <queue> 7 #include <set> 8 #include <stack> 9 #include <vector> 10 #include <deque> 11 #include <algorithm> 12 #include <cstdio> 13 #include <cmath> 14 #include <ctime> 15 #include <cstring> 16 #include <climits> 17 #include <cctype> 18 #include <cassert> 19 #include <functional> 20 #include <iterator> 21 #include <iomanip> 22 using namespace std; 23 //#pragma comment(linker,"/STACK:102400000,1024000") 24 25 #define sti set<int> 26 #define stpii set<pair<int, int> > 27 #define mpii map<int,int> 28 #define vi vector<int> 29 #define pii pair<int,int> 30 #define vpii vector<pair<int,int> > 31 #define rep(i, a, n) for (int i=a;i<n;++i) 32 #define per(i, a, n) for (int i=n-1;i>=a;--i) 33 #define clr clear 34 #define pb push_back 35 #define mp make_pair 36 #define fir first 37 #define sec second 38 #define all(x) (x).begin(),(x).end() 39 #define SZ(x) ((int)(x).size()) 40 #define lson l, mid, rt<<1 41 #define rson mid+1, r, rt<<1|1 42 43 typedef struct node_t { 44 int q, p; 45 46 node_t() {} 47 node_t(int q, int p): 48 q(q), p(p) {} 49 50 friend bool operator< (const node_t& a, const node_t& b) { 51 if (a.q == b.q) 52 return a.p < b.p; 53 return a.q < b.q; 54 } 55 56 } node_t; 57 58 const int maxn = 1e6+5; 59 int a[maxn]; 60 __int64 sum[maxn]; 61 node_t nd[maxn]; 62 int n, p, k; 63 int Q[maxn]; 64 __int64 ans; 65 66 void solve() { 67 int l = 1, r = 0; 68 __int64 tmp; 69 ans = -1; 70 71 rep(i, 0, n+1) { 72 while (l<=r && nd[i].p < nd[Q[r]].p) 73 --r; 74 Q[++r] = i; 75 while (l<=r && (nd[i].q - nd[Q[l]].q) > k) 76 ++l; 77 if (l >= r) 78 continue; 79 tmp = sum[nd[i].p] - sum[nd[Q[l]].p]; 80 ans = max(ans, tmp/p); 81 } 82 } 83 84 int main() { 85 ios::sync_with_stdio(false); 86 #ifndef ONLINE_JUDGE 87 freopen("data.in", "r", stdin); 88 freopen("data.out", "w", stdout); 89 #endif 90 91 int t; 92 93 scanf("%d", &t); 94 rep(tt, 1, t+1) { 95 scanf("%d %d %d", &n, &p, &k); 96 nd[0].p = nd[0].q = 0; 97 sum[0] = 0; 98 rep(i, 1, n+1) { 99 scanf("%d", &a[i]); 100 sum[i] = sum[i-1] + a[i]; 101 nd[i].p = i; 102 nd[i].q = sum[i] % p; 103 } 104 sort(nd, nd+n+1); 105 solve(); 106 printf("Case %d: %I64d ", tt, ans); 107 } 108 109 #ifndef ONLINE_JUDGE 110 printf("time = %d. ", (int)clock()); 111 #endif 112 113 return 0; 114 }