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  • codeforces895E. Eyes Closed

    题目链接

    codeforces895E. Eyes Closed

    题解

    线段树维护期望和
    写出修改两区间的相互影响
    就是一个区间修改

    emmm考试的代码过不去,这么松的spj都过不去Orz,每次和答案差0.001*ans左右...
    可能是我每次直接对区间暴力统计右区间的影响,然后直接打个修改区间tag....
    求解

    代码

    std:

    #include <iostream>
    #include <cstring>
    #include <cmath>
    #include <cstdio>
    #include <algorithm>
    #include <queue>
    #include <vector>
    #include <map>
    #include <complex>
    
    #define inf 0x3f3f3f3f
    #define eps 1e-10
    
    #define lc k << 1
    #define rc k << 1 | 1
    
    using namespace std;
    
    typedef long long ll;
    typedef pair<ll, int> P;
    ll p;
    double tmp[200005];
    
    struct node{
        double dat, tag1, tag2;
        int l, r;
    };
    struct seg{
        node d[800005];
        
        void pushup(int k){
            d[k].dat = d[lc].dat + d[rc].dat;
        }
        
        void build(int k, int l, int r){
            d[k].l = l; d[k].r = r; d[k].tag1 = 0; d[k].tag2 = 1;
            if(l == r){
                d[k].dat = tmp[l];
                return;
            }
            int mid = (l + r) >> 1;
            build(lc, l, mid);
            build(rc, mid + 1, r);
            pushup(k);
        }
        
        void add(int k, double x){
            double len = d[k].r - d[k].l + 1;
            d[k].dat = (d[k].dat + x * len);
            d[k].tag1 = (d[k].tag1 + x);
        }
        
        void mul(int k, double x){
            d[k].dat = d[k].dat * x;
            d[k].tag1 = d[k].tag1 * x;
            d[k].tag2 = d[k].tag2 * x;
        }
        
        void pushdown(int k){
            if(fabs(d[k].tag2 - 1) > eps){
                mul(lc, d[k].tag2);
                mul(rc, d[k].tag2);
                d[k].tag2 = 1;
            }
            if(fabs(d[k].tag1) > eps){
                add(lc, d[k].tag1);
                add(rc, d[k].tag1);
                d[k].tag1 = 0;
            }
        }
        
        void add(int k, int l, int r, double x){
            if(l <= d[k].l && d[k].r <= r){
                add(k, x); return;
            }
            pushdown(k);
            int mid = (d[k].l + d[k].r) >> 1;
            if(l <= mid) add(lc, l, r, x);
            if(r > mid) add(rc, l, r, x);
            pushup(k);
        }
        
        void mul(int k, int l, int r, double x){
            if(l <= d[k].l && d[k].r <= r){
                mul(k, x); return;
            }
            pushdown(k);
            int mid = (d[k].l + d[k].r) >> 1;
            if(l <= mid) mul(lc, l, r, x);
            if(r > mid) mul(rc, l, r, x);
            pushup(k);
        }
        
        double query(int k, int l, int r){
            if(l <= d[k].l && d[k].r <= r){
                return d[k].dat;
            }
            pushdown(k); double sum = 0;
            int mid = (d[k].l + d[k].r) >> 1;
            if(l <= mid) sum = (sum + query(lc, l, r));
            if(r > mid) sum = (sum + query(rc, l, r));
            return sum;
        }
        
    }Seg;
    
    int n, m;
    
    int main(){
    		
    	//freopen("random.in", "r", stdin);freopen("random.out", "w", stdout);
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; i ++) scanf("%lf", &tmp[i]);
        Seg.build(1, 1, n);
        while(m--){
            int opt, l1, r1, l2, r2; ll x;
            scanf("%d", &opt);
            if(opt == 1){
                scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
               	double d1 = Seg.query(1, l1, r1);
             	double d2 = Seg.query(1, l2, r2);
               	Seg.mul(1, l1, r1, double(r1 - l1) / double(r1 - l1 + 1));
             	Seg.mul(1, l2, r2, double(r2 - l2) / double(r2 - l2 + 1));
    			Seg.add(1, l1, r1, d2 / double(r1 - l1 + 1) / double(r2 - l2 + 1));
             	Seg.add(1, l2, r2, d1 / double(r2 - l2 + 1) / double(r1 - l1 + 1));         	  	
            }
            if(opt == 2){
                scanf("%d%d", &l1, &r1);
                printf("%.8lf
    ", Seg.query(1, l1, r1));
            }
        }
        return 0;
    }
    

    mycode

    #include<cmath> 
    #include<cstdio>
    #include<iostream>  
    #include<algorithm> 
    inline int read() { 	
    	int x = 0 ,f = 1; 
    	char c = getchar(); 
    	while(c > '9' || c < '0') c = getchar(); 
    	while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = getchar(); 
    	return x * f; 
    } 
    
    int n,m;  
    #define double long double
    const int maxn = 2000007; 
    int a[maxn]; 
    double t[maxn << 1] ;  
     double tag[maxn << 1]; 
    int tagl[maxn << 1],tagr[maxn << 1]; 
    
    void update(int x) { 
    	t[x] = t[x << 1] + t[x << 1 | 1];  
    } 
    
    double merge(double c1,double c2,int l,int r,int t1,int t2) { 
    	double k1 = 1.0 * r - l + 1, k2 = 1.0 * t2; 
    	return c1 * (k1 - 1) / k1 + c2 /t2 ; 
    } 
    
    void pushdown(int x,int l,int r) { 
    	int mid = l + r >> 1; 
    	
    	t[x << 1] = merge(t[x << 1] , tag[x] , l , mid , tagl[x] , tagr[x]); 
    	 
    	t[x << 1 | 1] = merge(t[x << 1 | 1] , tag[x] , mid + 1,r , tagl[x] , tagr[x]); 
    	
    	tag[x << 1] = tag[x << 1 | 1] = tag[x]; 
    	tagl[x << 1] = tagl[x << 1 | 1] = tagl[x]; 
    	tagr[x << 1] = tagr[x << 1 | 1] = tagr[x]; 
    	tagl[x] = tagr[x] = 0; tag[x] = 0.0; 
    } 
    
    void build(int x,int l,int r) { 
    	if(l == r) { t[x] = a[l]; return; } 
    	
    	int mid = l + r >> 1; 
    	
    	build(x << 1,l,mid); build(x << 1 | 1,mid + 1,r); 
    	update(x); 
    } 
    double Query(int x,int l,int r,int L,int R) { 
    	if(tag[x] != 0.0 && tagl[x] && tagr[x]) pushdown(x,l,r); 	
    	if(l >= L && r <= R)  return t[x];  
    	
    	double ret = 0; 
    	int mid = l + r >> 1; 
    	if(L <= mid) ret += Query(x << 1,l,mid,L,R); 
    	if(R > mid ) ret += Query(x << 1 | 1,mid + 1,r,L,R); 
    	return ret; 
    	//update(x); 
    } 
    void modify(int x,int l,int r,int L,int R,double CC,int t1,int t2) { 
    	
    	if(tag[x] != 0.0&& tagr[x] && tagl[x]) pushdown(x,l,r); 
    	if(l >= L && r <= R) { 
    		t[x] = merge(t[x],CC,l,r,t1,t2); 
    		if(l != r) { 
    			tag[x] = CC;  
    			tagl[x] = t1; 
    			tagr[x] = t2; 
    		} 
    		return ; 
    	} 
    	int mid = l + r >> 1; 
    	if(L <= mid) modify(x << 1,l,mid,L,R,CC, t1,t2); 
    	if(R > mid ) modify(x << 1 | 1,mid + 1,r,L,R, CC,t1,t2); 
    	update(x); 
    } 
    int main() { 
    	//freopen("random.in","r",stdin);  freopen("random.out","w",stdout); 
    	n = read(),m = read(); 
    	for(int i = 1;i <= n;++ i) a[i] = read();  
    		
    	build(1,1,n); 
    	for(int type,l,r,l1,r1,i = 1;i <= m;++ i) { 
    		type = read(); 
    		if(type == 1) { 
    			l = read(),r = read(); 
    			l1 = read();r1 = read(); 
    			double q1 = Query(1,1,n,l,r); 
    			double q2 = Query(1,1,n,l1,r1); 
    			modify(1,1,n,l,r,q2 , r - l + 1 , r1 - l1 + 1); 
    			modify(1,1,n,l1,r1,q1 , r1 - l1 + 1 , r - l + 1); 
    		} 
    		else { 
    			l = read(),r = read(); 
    			//printf("%llf
    ",Query(1,1,n,l,r)); 
    			std::cout << Query(1,1,n,l,r) << std::endl; 
    		} 
    	} 
    	return 0; 
    } 
    /* 
    4 4
    1 1 2 2
    1 2 2 3 3
    2 1 2
    1 1 2 3 4
    2 1 2
    */ 
    
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  • 原文地址:https://www.cnblogs.com/sssy/p/9538562.html
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