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  • CF438D The Child and Sequence(线段树)

    区间驱魔?看你大于模数吗,大就上,否则回家。
    会不会(T)?每次驱魔至少变(frac{1}{2}),所以(log),不怂
    注意到叶子才更新

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cstdlib>
    #include <algorithm>
    #include <cmath>
    #define R(a,b,c) for(register int a = (b); a <= (c); ++a)
    #define nR(a,b,c) for(register int a = (b); a >= (c); --a)
    #define Fill(a,b) memset(a, b, sizeof(a))
    #define Swap(a,b) ((a) ^= (b) ^= (a) ^= (b))
    #define QWQ
    #ifdef QWQ
    #define D_e_Line printf("
    ---------------
    ")
    #define D_e(x) cout << (#x) << " : " << x << "
    "
    #define Pause() system("pause")
    #define FileOpen() freopen("in.txt", "r", stdin)
    #define FileSave() freopen("out.txt", "w", stdout)
    #define TIME() fprintf(stderr, "
    TIME : %.3lfms
    ", clock() * 1000.0 / CLOCKS_PER_SEC)
    #else
    #define D_e_Line ;
    #define D_e(x) ;
    #define Pause() ;
    #define FileOpen() ;
    #define FileSave() ;
    #define TIME() ;
    #endif
    struct ios {
    	template<typename ATP> inline ios& operator >> (ATP &x) {
    		x = 0; int f = 1; char c;
    		for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-') f = -1;
    		while(c >= '0' && c <='9') x = x * 10 + (c ^ '0'), c = getchar();
    		x *= f;
    		return *this;
    	}
    }io;
    using namespace std;
    template<typename ATP> inline ATP Max(ATP a, ATP b) {
    	return a > b ? a : b;
    }
    template<typename ATP> inline ATP Min(ATP a, ATP b) {
    	return a < b ? a : b;
    }
    template<typename ATP> inline ATP Abs(ATP a) {
    	return a < 0 ? -a : a;
    }
    
    const int N = 1e5 + 7;
    
    int n;
    
    long long t[N << 2];
    int mx[N << 2];
    
    #define ls rt << 1
    #define rs rt << 1 | 1
    #define lson rt << 1, l, mid
    #define rson rt << 1 | 1, mid + 1, r
    inline void Pushup(int &rt) {
    	t[rt] = t[ls] + t[rs];
    	mx[rt] = Max(mx[ls], mx[rs]);
    }
    inline void Build(int rt, int l, int r) {
    	if(l == r){
    		io >> t[rt];
    		mx[rt] = t[rt];
    		return;
    	}
    	int mid = (l + r) >> 1;
    	Build(lson), Build(rson);
    	Pushup(rt);
    }
    inline void Updata(int rt, int l, int r, int L, int R, int mod) {
    	if(mx[rt] < mod) return;
    	if(l == r){
    		t[rt] %= mod;
    		mx[rt] %= mod;
    		return;
    	}
    	int mid = (l + r) >> 1;
    	if(L <= mid) Updata(lson, L, R, mod);
    	if(R > mid) Updata(rson, L, R, mod);
    	Pushup(rt);
    }
    inline long long Query(int rt, int l, int r, int L, int R) {
    	if(L <= l && r <= R) return t[rt];
    	int mid = (l + r) >> 1;
    	long long sum = 0;
    	if(L <= mid) sum += Query(lson, L, R);
    	if(R > mid) sum += Query(rson, L, R);
    	return sum;
    }
    inline int QueryMax(int rt, int l, int r, int L, int R) {
    	if(L <= l && r <= R) return mx[rt];
    	int mid = (l + r) >> 1;
    	int maxx = -1e9;
    	if(L <= mid) maxx = Max(maxx, QueryMax(lson, L, R));
    	if(R > mid) maxx = Max(maxx, QueryMax(rson, L, R));
    	return maxx;
    }
    inline void Modify(int rt, int l, int r, int x, int w) {
    	if(l == r){
    		t[rt] = w;
    		mx[rt] = w;
    		return;
    	}
    	int mid = (l + r) >> 1;
    	if(x <= mid)
    		Modify(lson, x, w);
    	else
    		Modify(rson, x, w);
    	Pushup(rt);
    }
    
    int main() {
    //freopen("mod.in", "r", stdin);
    //freopen("mod.out", "w", stdout);
    	int m;
    	io >> n >> m;
    	
    	Build(1, 1, n);
    
    	while(m--){
    		int opt;
    		io >> opt;
    		if(opt == 1){
    			int l, r;
    			io >> l >> r;
    			printf("%lld
    ", Query(1, 1, n, l, r));
    		}
    		else if(opt == 2){
    			int l, r, mod;
    			io >> l >> r >> mod;
    //			if(QueryMax(1, 1, n, l, r) < mod) continue;
    //			R(i,l,r){
    //				Updata(1, 1, n, i, mod);
    //			}
    			Updata(1, 1, n, l, r, mod);
    		}
    		else{
    			int x, w;
    			io >> x >> w;
    			Modify(1, 1, n, x, w);
    		}
    	}
    	return 0;
    }
    /*
    5 999
    1 2 3 4 5 
    2 3 5 4 
    3 3 5 
    1 2 5 
    2 1 3 3 
    1 1 3 
    */
    

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  • 原文地址:https://www.cnblogs.com/bingoyes/p/11737301.html
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