Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37270 Accepted Submission(s): 15723
1 #include<stdio.h> 2 #define maxn 50005 3 int aa[maxn]; 4 struct node 5 { 6 int lef,rig,sum; 7 int mid(){ 8 return (lef +(rig-lef)/2); 9 } 10 }; 11 node tree[maxn<<2]; 12 void build(int left ,int right ,int p) 13 { 14 tree[p].lef=left; //p代表当前节点位置 15 tree[p].rig=right; 16 tree[p].sum=0; 17 if(left==right) 18 { 19 tree[p].sum=aa[left]; 20 return ; 21 } 22 int mid=tree[p].mid(); 23 build(left,mid,p*2); 24 build(mid+1,right,p*2+1); 25 tree[p].sum=tree[p*2].sum + tree[p*2+1].sum; 26 } 27 28 void updata(int pos ,int p ,int val) 29 { 30 tree[p].sum+= val; //¸更新数据 31 int mid=tree[p].mid(); 32 if(tree[p].lef==pos&&tree[p].rig==pos) return ; 33 if(pos<=mid) updata(pos,p<<1,val); 34 else 35 if(pos>mid) updata(pos,p<<1|1,val); 36 } 37 38 int query(int be,int en,int p) 39 { 40 int res=0; 41 if(be<=tree[p].lef&&tree[p].rig<=en) 42 return tree[p].sum ; 43 int mid=tree[p].mid(); 44 if(be<=mid) res+=query(be,en,2*p); 45 if(en>mid) res+=query(be,en,2*p+1); 46 return res; 47 } 48 int main() 49 { 50 int tt,nn,i,j,sa,sb; 51 char str[6]; 52 scanf("%d",&tt); 53 for(j=1;j<=tt;j++) 54 { 55 scanf("%d",&nn); 56 for(i=1;i<=nn;i++) 57 scanf("%d",&aa[i]); 58 build(1,nn,1); 59 printf("Case %d:\n",j); 60 while(1) 61 { 62 scanf("%s",str); 63 if(str[0]=='E') break; 64 else 65 { 66 scanf("%d%d",&sa,&sb); 67 if(str[0]=='A') updata(sa,1,sb); 68 else if(str[0]=='S') updata(sa,1,-sb); 69 else printf("%d\n",query(sa,sb,1)); 70 } 71 } 72 } 73 return 0; 74 }