题意
题目链接
单点修改，区间mod，区间和
Sol
如果x > mod ，那么 x % mod < x / 2
证明：
即得易见平凡， 仿照上例显然， 留作习题答案略， 读者自证不难。 反之亦然同理， 推论自然成立， 略去过程Q.E.D.， 由上可知证毕。
然后维护个最大值就做完了。。
复杂度不知道是一个log还是两个log，大概是两个吧(线段树一个 最多改log次。)
代码语言：javascript复制#include<bits/stdc .h>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long
#define LL long long
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 1e6 10, mod = 1e9 7, INF = 1e9 10;
const double eps = 1e-9;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline LL add(A x, B y) {if(x y < 0) return x y mod; return x y >= mod ? x y - mod : x y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x y < 0) x = x y mod; else x = (x y >= mod ? x y - mod : x y);}
template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod mod) % mod;}
template <typename A> inline void debug(A a){cout << a << 'n';}
template <typename A> inline LL sqr(A x){return 1ll * x * x;}
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 c - '0', c = getchar();
return x * f;
}
int N, M, a[MAXN];
#define ls k << 1
#define rs k << 1 | 1
LL sum[MAXN];
int mx[MAXN];
void update(int k) {
sum[k] = sum[ls] sum[rs];
mx[k] = max(mx[ls], mx[rs]);
}
void Build(int k, int ll, int rr) {
if(ll == rr) {sum[k] = mx[k] = read(); return ;}
int mid = ll rr >> 1;
Build(ls, ll, mid); Build(rs, mid 1, rr);
update(k);
}
LL Query(int k, int l, int r, int ql, int qr) {
if(ql <= l && r <= qr) return sum[k];
int mid = l r >> 1;
if(ql > mid) return Query(rs, mid 1, r, ql, qr);
else if(qr <= mid) return Query(ls, l, mid, ql, qr);
else return Query(ls, l, mid, ql, qr) Query(rs, mid 1, r, ql, qr);
}
void Modify(int k, int l, int r, int p, int v) {
if(l == r) {sum[k] = mx[k] = v; return ;}
int mid = l r >> 1;
if(p <= mid) Modify(ls, l, mid, p, v);
else Modify(rs, mid 1, r, p, v);
update(k);
}
void Mod(int k, int l, int r, int ql, int qr, int x) {
if(mx[k] < x) return ;
if(l == r) {sum[k] = mx[k] % x; mx[k] %= x; return ;}
int mid = l r >> 1;
if(ql <= mid) Mod(ls, l, mid, ql, qr, x);
if(qr > mid) Mod(rs, mid 1, r, ql, qr, x);
update(k);
}
signed main() {
N = read(); M = read();
Build(1, 1, N);
while(M--) {
int opt = read();
if(opt == 1) {
int l = read(), r = read();
cout << Query(1, 1, N, l, r) << 'n';
} else if(opt == 2) {
int l = read(), r = read(), x = read();
Mod(1, 1, N, l, r, x);
} else {
int k = read(), x = read();
Modify(1, 1, N, k, x);
}
}
return 0;
}
/*
*/
