poj 3468 A Simple Problem with Integers

线段树模板题

#include <stdio.h>
#define L(t) ((t) << 1)
#define R(t) ((t) << 1 | 1)
#define MAXN 100010
struct SegTree
{
    int l,r;
    long long add, sum;    
    int getMid(){
        return ( l + r) >> 1;
    }
    int getDis(){
        return r - l + 1;
    }
}       tree[MAXN << 2];

int arr[MAXN];

void build(int left, int right, int t){           //递归构造
    tree[t].l = left;
    tree[t].r = right;
    tree[t].add = 0;
    if(left == right){
        tree[t].sum = arr[left];                      //叶结点是数列中的一个数
        return;
    }
    int mid = tree[t].getMid();
    build(left, mid, L(t));
    build(mid + 1, right, R(t));
    tree[t].sum = tree[L(t)].sum + tree[R(t)].sum;          //更新sum
}
void update(int left, int right, int a, int t){            
    if( left <= tree[t].l && right >= tree[t].r ){                
        tree[t].add += a;            //若当前子树被目标区间覆盖 更新子树的sum和增量（在本次查询中，增量下行到此为之）
        tree[t].sum += a * tree[t].getDis();    
        return;
    }
    if( tree[t].add ){                //向子结点传递增量，并更新其sum,最后清空自己的增量
        tree[L(t)].sum += tree[L(t)].getDis() * tree[t].add;
        tree[R(t)].sum += tree[R(t)].getDis() * tree[t].add;
        tree[L(t)].add += tree[t].add;
        tree[R(t)].add += tree[t].add;
        tree[t].add = 0;        
    }
    int mid = tree[t].getMid();
    if(right <= mid ){
        update(left, right, a, L(t));       //目标区间仅在左子树上
    } else if (left > mid ){
        update(left, right, a, R(t));           //目标区间仅在右子树上
    } else {
        update(left, mid, a, L(t));                          //目标区间同时在左右子树上
        update(mid + 1, right, a, R(t));
    }    
    tree[t].sum = tree[L(t)].sum + tree[R(t)].sum ;    //更新父结点的sum
}
long long query(int left, int right, int t){
    if(left <= tree[t].l && right >= tree[t].r ){    
        return tree[t].sum;
    }
    if( tree[t].add ){               //这一段和update函数一样，是一个pushDown
        tree[L(t)].sum += tree[L(t)].getDis() * tree[t].add;
        tree[R(t)].sum += tree[R(t)].getDis() * tree[t].add;
        tree[L(t)].add += tree[t].add;
        tree[R(t)].add += tree[t].add;
        tree[t].add = 0;
    }
    int mid = tree[t].getMid();
    if(right <= mid ){
        return query(left, right, L(t));    
    } else if( left > mid ){
        return query(left, right, R(t));
    }  else {
        return query(left, mid ,L(t)) + 
        query(mid + 1, right, R(t));
    }
}
int main(){
    char op[5];
    int n, q;
    scanf("%d%d",&n, &q);
    for(int i = 1; i <= n; ++i){
        scanf("%d",&arr[i]);
    }
    build(1, n, 1);       
	int a,b,c;
    for(int i = 0; i < q; ++i){
        scanf("%s ",&op);

        if(op[0] == 'Q'){
           
            scanf("%d %d",&a, &b);
            long long sum = query(a, b, 1);
            printf("%lld\n",sum);
        } else {
           
            scanf("%d %d %d",&a, &b, &c);
            update(a, b, c, 1);
        }
    }
    return 0;
}

此题的数组数组解法很巧妙，可见对于前缀和问题树状数组也是很有效的
见   http://kenby.iteye.com/blog/962159


PS：读入数据不要用long long   会WA的。。。被坑了很久

#include<stdio.h>  
#define MAXC 100010  
long long  C[MAXC], D[MAXC], DQ[MAXC];  
int A[MAXC];
//////////////////树状数组模板  
int lowbit(int t)                
{  
    return t & (t ^ (t - 1));      //计算最小幂  
	
}  
int add(int i,long long v,int n)  
{  
	if(n==1)
  {  
	  while(i<=MAXC)      //取与C[i]有关的值添上增加的值  
    {  
         C[i]+=v;  
         i+=lowbit(i);  
    }  
	return 0; 
	}
	if(n==2)
  {  
	  while(i<=MAXC)      
    {  
         D[i]+=v;  
         i+=lowbit(i);  
    }  
	return 0; 
	}
	if(n==3)
  {  
	  while(i<=MAXC)      
    {  
         DQ[i]+=v;  
         i+=lowbit(i);  
    }  
	return 0; 
	}
	return 0;
}  
long long sum(int i,int n)  
{  
    long long s=0;  
	if(n==1)
	{
    while(i>0)                //取与C[i]有关的值求和  
    {  
         s+=C[i];  
         i-=lowbit(i);  
    }  
    return s;  
	}
	if(n==2)
	{
    while(i>0)                
    {  
         s+=D[i];  
         i-=lowbit(i);  
    }  
    return s;  
	}
	if(n==3)
	{
    while(i>0)               
    {  
         s+=DQ[i];  
         i-=lowbit(i);  
    }  
    return s;  
	}
	return 0;
}  
//////////////////树状数组模板  
int main(){  



    char op[5];  
    int n, q;  
    scanf("%d%d",&n, &q);  

    for(int i = 1; i <= n; ++i){  
        scanf("%d",&A[i]);  
	    add(i,A[i],1);
		
    }  
    
    int a,b,c;  
    for(int i = 0; i < q; ++i){  
        scanf("%s ",&op);  
  
        if(op[0] == 'Q'){  
             
            scanf("%d %d",&a, &b);  
		
			long long ss =( sum(b,1) + (b+1)*sum(b,2) - sum(b,3) ) - ( sum(a-1,1) + (a)*sum(a-1,2) - sum(a-1,3) );  
            printf("%lld\n",ss);  
        } else {  
             
            scanf("%d %d %d",&a, &b, &c);  
           
			
			add(a,c,2);
			add(b+1,-c,2);
			add(a,c*a,3);
		    add(b+1,-c*(b+1),3);
        }  
    }  
    return 0;  
}