cf 442 div2 F. Ann and Books(莫队算法)

cf 442 div2 F. Ann and Books(莫队算法)

题意：

(给出n和k,和a_i,sum_i表示前i个数的和,有q个查询[l,r])
每次查询区间([l,r]内有多少对(i,j)满足l <= i <= j <= r 且 sum[j] - sum[i-1] = k)

思路:

区间左右端点的挪动对答案的贡献符合加减性质，直接用莫队算法即可
复杂度(O(n * sqrt(n) * log(maxsum))) 过高
考虑先离散化预处理出所有位置 将(log)去掉

#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int N = 2e5 + 10;
struct Q{
    int l,r,bl,id;
    Q(){};
    bool operator<(const Q&rhs){
         if(bl == rhs.bl) return r < rhs.r;
         return bl < rhs.bl;
    }
}qr[N];
int n,k;
LL ans[N],value[N];
int x[N],y[N],z[N],cnt[N * 3],type[N];
vector<LL> se;
int main(){

    while(cin>>n>>k){
        memset(cnt, 0, sizeof(cnt));
        se.clear();
        for(int i = 1;i <= n;i++) scanf("%d",&type[i]);
        for(int i = 1;i <= n;i++){
            scanf("%d",&value[i]);
            if(type[i] == 1) value[i] += value[i-1];
            else value[i] = value[i-1] - value[i];
        }
        for(int i = 0;i <= n;i++) {
            se.push_back(value[i]);
            se.push_back(value[i] + k);
            se.push_back(value[i] - k);
        }
        sort(se.begin(), se.end());
        se.erase(unique(se.begin(),se.end()),se.end());
        for(int i = 0;i <= n;i++){
            x[i] = lower_bound(se.begin(),se.end(),value[i]) - se.begin();
            y[i] = lower_bound(se.begin(),se.end(),value[i] + k) - se.begin();
            z[i] = lower_bound(se.begin(),se.end(),value[i] - k) - se.begin();
        }
        int block_size = sqrt(n + 0.5);
        int q;
        cin>>q;
        for(int i = 0;i < q;i++){
            scanf("%d%d",&qr[i].l,&qr[i].r);
            qr[i].id = i;
            qr[i].l--;
            qr[i].bl = qr[i].l / block_size;
        }
        sort(qr, qr + q);
        int L = 0,R = -1;
        LL res = 0;
        for(int i = 0;i < q;i++){
                while(qr[i].l > L) {
                    cnt[x[L]]--;
                    res -= cnt[y[L++]];
                }
                while(qr[i].l < L) {
                    res += cnt[y[--L]];
                    cnt[x[L]]++;
                }
                while(qr[i].r > R){
                    res += cnt[z[++R]];
                    cnt[x[R]]++;
                }
                while(qr[i].r < R) {
                    cnt[x[R]]--;
                    res -= cnt[z[R--]];
                }
            ans[qr[i].id] = res;
        }
        for(int i = 0;i < q;i++) printf("%lld
",ans[i]);
    }
    return 0;
}