POJ 3468 A Simple Problem with Integers（线段树 单点更新+区间求和 ）

  1 //POJ 3468 A Simple Problem with Integers
  2 //单点更新 区间求和 
  3 #include <cstdio>
  4 #include <iostream>
  5 #include <algorithm>
  6 using namespace std;
  7 
  8 typedef long long LL;
  9 const int N=100000+10;
 10 LL ans,n,m;
 11 
 12 struct Tree
 13 {
 14     LL l,r;
 15     LL sum,add;
 16 };
 17 Tree tree[4*N];
 18 
 19 void pushup(LL x) //向上更新 
 20 {
 21     LL tmp=x<<1;
 22     tree[x].sum=tree[tmp].sum+tree[tmp+1].sum;
 23 }
 24 
 25 void pushdown(LL x) //向下更新 
 26 {
 27     LL tmp=x<<1;
 28     tree[tmp].add+=tree[x].add;
 29     tree[tmp+1].add+=tree[x].add;
 30     tree[tmp].sum+=tree[x].add*(tree[tmp].r-tree[tmp].l+1);
 31     tree[tmp+1].sum+=tree[x].add*(tree[tmp+1].r-tree[tmp+1].l+1);
 32     tree[x].add=0;
 33 }
 34 
 35 void build(LL l,LL r,LL x)
 36 {
 37     tree[x].l=l;
 38     tree[x].r=r;
 39     tree[x].add=0;
 40     if(l==r)
 41     {
 42         scanf("%lld",&tree[x].sum);
 43         return ;
 44     }    
 45     LL tmp=x<<1;
 46     LL mid=(l+r)>>1;
 47     build(l,mid,tmp);
 48     build(mid+1,r,tmp+1);
 49     pushup(x);
 50 }
 51 
 52 void update(LL l,LL r,LL c,LL x)
 53 {
 54     if(r<tree[x].l||l>tree[x].r) return ;
 55     if(l<=tree[x].l&&r>=tree[x].r)
 56     {
 57         tree[x].add+=c;
 58         tree[x].sum+=c*(tree[x].r-tree[x].l+1);
 59         return ;    
 60     }
 61     if(tree[x].add) pushdown(x);
 62     LL tmp=x<<1;
 63     update(l,r,c,tmp);
 64     update(l,r,c,tmp+1);
 65     pushup(x);
 66 }
 67 
 68 void query(LL l,LL r,LL x)
 69 {
 70     if(r<tree[x].l||l>tree[x].r) return ;
 71     if(l<=tree[x].l&&r>=tree[x].r)
 72     {
 73         ans+=tree[x].sum;
 74         return ;
 75     }
 76     if(tree[x].add) pushdown(x);    
 77     LL tmp=x<<1;
 78     LL mid=(tree[x].l+tree[x].r)>>1;
 79     if(r<=mid) query(l,r,tmp);
 80     else if(l>mid) query(l,r,tmp+1);
 81     else
 82     {
 83         query(l,mid,tmp);
 84         query(mid+1,r,tmp+1);    
 85     }
 86 }
 87 
 88 
 89 int main(){
 90     char op[5];
 91     LL A,B,C;
 92     scanf("%lld %lld",&n,&m);
 93     build(1,n,1);
 94     for(int i=1;i<=m;i++){
 95         scanf("%s",op);
 96         if(op[0]==‘Q‘){
 97             ans=0;
 98             scanf("%lld%lld",&A,&B);
 99             query(A,B,1);
100             printf("%lld\n",ans);
101         }
102         else if(op[0]==‘C‘){
103             scanf("%lld%lld%lld",&A,&B,&C);
104             update(A,B,C,1);
105         }
106     }
107     return 0;
108 }