poj3468——A Simple Problem with Integers

线段树之整段更新操作

思路：为提高效率，每次有insert操作的时候，如果刚好一个区间的数都需加ad，那么，将此区间的st[i].ad赋为ad。查找过程，如果经过此区间的话，那么，需做ans+=st[i].ad*(r-l+1)，此处的r,l是指要查的那个区间！

#include<stdio.h>#include<string.h>#define max 100005struct node{int l,r;__int64 add,sum;}st[max*4];int n,m;__int64 num[max],ans;void create(int l,int r,int index){st[index].l=l;st[index].r=r;st[index].add =0;st[index].sum =0;if(st[index].r==st[index].l)return;int mid=(st[index].l+st[index].r)>>1;create(l,mid,index*2);create(mid+1,r,index*2+1);}void build(int p,int t,int i){st[i].sum +=t;if(st[i].r==st[i].l)return ;int mid=(st[i].r+st[i].l)>>1;if(p<=mid)build(p,t,i*2);elsebuild(p,t,i*2+1);}void Insert(int l,int r,int ad,int i){if(l==st[i].l&&r==st[i].r){st[i].add+=ad;return ;}else {st[i].sum +=ad*(r-l+1);}int mid=(st[i].l+st[i].r)>>1;if(r<=mid)Insert(l,r,ad,i*2);else if(l>mid)Insert(l,r,ad,i*2+1);else{Insert(l,mid,ad,i*2);Insert (mid+1,r,ad,i*2+1);}}void Q(int l,int r,int i){if(st[i].l==l&&st[i].r==r){ans+=st[i].sum;ans+=st[i].add*(st[i].r-st[i].l+1);return;}else ans+=st[i].add*(r-l+1);int mid=(st[i].l+st[i].r)>>1;if(r<=mid)Q(l,r,i*2);else if(l>mid)Q(l,r,i*2+1);else{Q(l,mid,i*2);Q(mid+1,r,i*2+1);}}int main(){char c;int i,x,y,z;scanf("%d%d",&n,&m);for(i=1;i<=n;i++)scanf("%I64d",&num[i]);create(1,n,1);for(i=1;i<=n;i++)build(i,num[i],1);for(i=0;i<m;i++){getchar();scanf("%c",&c);if(c=='C'){scanf("%d%d%d",&x,&y,&z);Insert(x,y,z,1);}else if(c=='Q'){scanf("%d%d",&x,&y);ans=0;Q(x,y,1);printf("%I64d/n",ans);}}return 0;}