P1438 无聊的数列

P1438 无聊的数列

链接

分析：

　　等差数列可加，首项相加，公差相加。

代码：

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cmath>
#include<cctype>
#include<set>
#include<queue>
#include<vector>
#include<map>
#include<bitset>
using namespace std;
typedef long long LL;

inline int read() {
    int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}

const int N = 1e5 + 10;
int A[N << 2], D[N << 2], ori[N];

void pushdown(int rt,int l,int r) {
    int mid = (l + r) >> 1;
    A[rt << 1] += A[rt];
    A[rt << 1 | 1] += A[rt] + (mid - l + 1) * D[rt]; 
    D[rt << 1] += D[rt];
    D[rt << 1 | 1] += D[rt];
    A[rt] = D[rt] = 0;
}
void update(int l,int r,int rt,int L,int R,int a,int d) {
    if (L <= l && r <= R) {
        A[rt] += a + (l - L) * d; D[rt] += d; return ;
    }
    if (A[rt] || D[rt]) pushdown(rt, l, r);
    int mid = (l + r) >> 1;
    if (L <= mid) update(l, mid, rt << 1, L, R, a, d);
    if (R > mid) update(mid + 1, r, rt << 1 | 1, L, R, a, d);
}
int query(int l,int r,int rt,int p) {
    if (l == r) return ori[l] + A[rt];
    if (A[rt] || D[rt]) pushdown(rt, l, r);
    int mid = (l + r) >> 1;
    if (p <= mid) return query(l, mid, rt << 1, p);
    else return query(mid + 1, r, rt << 1 | 1, p);
}
int main() {
    int n = read(), m = read();
    for (int i = 1; i <= n; ++i) ori[i] = read();
    while (m --) {
        if (read() == 1) {
            int l = read(), r = read(), a = read(), d = read();
            update(1, n, 1, l, r, a, d);
        }
        else {
            int x = read();
            cout << query(1, n, 1, x) << "\n";
        }
    }
    return 0;
}