HDU5286 wyh2000 and sequence【分块 均摊复杂度】

HDU5286 wyh2000 and sequence

题意：

给出长为\(N\)的序列\(A_1,A_2,A_3,\cdots,A_n\)，\(q\)次询问，每次询问给出区间\([L,R]\)，假设区间内出现过的数为\(C_1,C_2,\cdots,C_k\)，出现的次数分别为\(B_1,B_2,\cdots,B_k\)，输出\(\sum_{i=1}^{k}C_i^{B_i} % 1000000007\)，要求在线

题解：

如果不要求在线，直接用莫队很快就能解决，现在要求在线，还是考虑把序列分块，预处理出所有连续的块的答案，和到第\(i\)块为止，各个数出现的次数的前缀和
对于每次查询，如果在一个块内，就直接暴力查询
否则对于在两端非整块的元素，记录元素大小和出现次数，将这些元素带来的贡献计算一下即可

有点卡常

//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MOD = 1e9+7;
const int MAXN = 5e4+7;
const int SQT = 240;
int n,q,m,sqt,A[MAXN],pre[SQT][MAXN],pos[MAXN],belong[MAXN],l[SQT],r[SQT],la;
int ans[SQT][SQT],num[MAXN];
vector<int> vec,cnt,pw[MAXN];
void divide(){
    sqt = pow(n,0.5);
    m = n / sqt + (n%sqt?1:0);
    for(int i = 1; i <= n; i++) belong[i] = (i-1) / sqt + 1;
    for(int i = 1; i <= m; i++) l[i] = (i-1) * sqt + 1, r[i] = i * sqt;
    r[m] = n;
}
void gao(){
    int L, R;
    scanf("%d %d",&L,&R);
    L = (L^la)%n+1; R = (R^la)%n+1;
    if(L>R) L ^= R ^= L ^= R;
    int lp = belong[L], rp = belong[R];
    if(lp==rp){
        int ret = 0;
        for(int i = L; i <= R; i++) num[pos[i]] = 0;
        for(int i = L; i <= R; i++){
            ret = ret - pw[pos[i]][num[pos[i]]];
            ret = ret + pw[pos[i]][++num[pos[i]]];
            if(ret<0) ret += MOD;
            if(ret>=MOD) ret -= MOD;
        }
        printf("%d\n",la=ret);
    }
    else{
        vector<int> arr;
        for(int i = L; i <= r[lp]; i++){
            num[pos[i]] = 0;
            arr.push_back(pos[i]);
        }
        for(int i = l[rp]; i <= R; i++){
            num[pos[i]] = 0;
            arr.push_back(pos[i]);
        }
        sort(arr.begin(),arr.end()); arr.erase(unique(arr.begin(),arr.end()),arr.end());
        for(int i = L; i <= r[lp]; i++) num[pos[i]]++;
        for(int i = l[rp]; i <= R; i++) num[pos[i]]++;
        lp++, rp--;
        int ret = ans[lp][rp];
        for(int i = 0; i < (int)arr.size(); i++){
            ret = ret - pw[arr[i]][pre[rp][arr[i]]-pre[lp-1][arr[i]]];
            ret = ret + pw[arr[i]][pre[rp][arr[i]]-pre[lp-1][arr[i]]+num[arr[i]]];
            if(ret<0) ret += MOD;
            if(ret>=MOD) ret -= MOD;            
        }
        printf("%d\n",la=ret);
    }
}
void solve(){
    scanf("%d %d",&n,&q);
    vec.clear();
    for(int i = 1; i <= n; i++){
        scanf("%d",&A[i]);
        vec.push_back(A[i]);
    }
    sort(vec.begin(),vec.end()); vec.erase(unique(vec.begin(),vec.end()),vec.end());
    for(int i = 1; i <= n; i++) pos[i] = lower_bound(vec.begin(),vec.end(),A[i]) - vec.begin();
    divide();
    for(int i = 1; i <= m; i++) for(int j = 0; j < (int)vec.size(); j++) pre[i][j] = 0;
    for(int i = 1; i <= n; i++) pre[belong[i]][pos[i]]++;
    for(int i = 1; i <= m; i++) for(int j = 0; j < (int)vec.size(); j++) pre[i][j] += pre[i-1][j];
    cnt.resize(vec.size());
    for(int i = 0; i < (int)vec.size(); i++) cnt[i] = 0;
    for(int i = 1; i <= n; i++) cnt[pos[i]]++;
    for(int i = 0; i < (int)vec.size(); i++){
        pw[i].resize(cnt[i]+1);
        pw[i][0] = 1; for(int j = 1; j <= cnt[i]; j++) pw[i][j] = 1ll * pw[i][j-1] * vec[i] % MOD;
        pw[i][0] = 0;
    }
    for(int i = 1; i <= m; i++){
        memset(num,0,sizeof(num));
        int p = l[i] - 1, ret = 0;
        for(int j = i; j <= m; j++){
            while(p<r[j]){
                p++;
                ret -= pw[pos[p]][num[pos[p]]];
                ret += pw[pos[p]][++num[pos[p]]];
                if(ret<0) ret += MOD;
                if(ret>=MOD) ret -= MOD;
            }
            ans[i][j] = ret;
        }
    }
    la = 0;
    while(q--) gao();
    for(int i = 0; i < (int)vec.size(); i++) pw[i].clear();
}
int main(){
    int T;
    for(scanf("%d",&T); T; T--) solve();
    return 0;
}