poj3468 A Simple Problem with Integers 线段树lazy标签

Description

You have N integers, A₁, A₂, ... , A_N. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A₁, A₂, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915

其实最开始我调了很多次，并且这个程序过不了，因为我懒得改成int64了，留着自己以后看看

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;struct segtree{long long l,r,v;}tree[500005];long long ans,add[500005];void build(long long l,long long r,long long  root){long long  m=(l+r)/2;tree[root].l=l;tree[root].r=r;tree[root].v=0;if(l==r)return;build(l,m,2*root);build(m+1,r,2*root+1);}void adjust(long long root,long long m){if(add[root]){add[root*2]+=add[root];add[root*2+1]+=add[root];tree[root*2].v+=add[root]*(m-(m/2));tree[root*2+1].v+=add[root]*(m/2);add[root]=0; }}void insert(long long v,long long l,long long  root){long long  m;if(tree[root].l==tree[root].r&&tree[root].l==l){tree[root].v+=v;return;}m=(tree[root].l+tree[root].r)/2;if(l<=m)insert(v,l,2*root);else insert(v,l,2*root+1);tree[root].v+=v;}void query(long long l,long long r,long long  root){long long  m=(tree[root].l+tree[root].r)/2;adjust(root,tree[root].r-tree[root].l+1);//每次查找时，把lazy标签处理了 if(tree[root].l==l&&tree[root].r==r){ans+=tree[root].v;return;}if(m>=r)query(l,r,2*root);else if(m<l)query(l,r,2*root+1);else{query(l,m,2*root);query(m+1,r,2*root+1);}}void update(long long c,long long l,long long r,long long root){if(tree[root].l==l&&tree[root].r==r){add[root]+=c;tree[root].v+=c*(r-l+1);//区间和加这么多 return;}if(tree[root].l==tree[root].r)return;adjust(root,tree[root].r-tree[root].l+1);long long m=(tree[root].l+tree[root].r)/2;if(r<=m)update(c,l,r,root*2);else if(l>m)update(c,l,r,root*2+1);else{update(c,l,m,root*2);update(c,m+1,r,root*2+1);}tree[root].v+=tree[root*2].v+tree[root*2+1].v;}int main(){long long  i,n,x,t,j;cin>>n>>t;build(1,n,1);for(i=1;i<=n;i++){cin>>j;insert(j,i,1);;}char s;while(t--){cin>>s;long long a,b,c;if(s=='C'){cin>>a>>b>>c;update(c,a,b,1);}else {cin>>a>>b;ans=0;query(a,b,1);cout<<ans<<endl;}}return 0;}

这是ac代码

#include<cstring>#include<cstdio>#include<iostream>#include<algorithm>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define LL long longconst int maxn=111111;LL add[maxn<<2],sum[maxn<<2];void pushup(int rt){sum[rt]=sum[rt<<1]+sum[rt<<1|1];}void build(int l,int r,int rt){add[rt]=0;if(l==r){scanf("%lld",&sum[rt]);return;}int m=(l+r)>>1;build(lson);build(rson);pushup(rt);}void pushdown(int rt,int m){if(add[rt]){add[rt<<1]+=add[rt];add[rt<<1|1]+=add[rt];sum[rt<<1]+=add[rt]*(m-(m>>1));sum[rt<<1|1]+=add[rt]*(m>>1);add[rt]=0;}}void update(int x,int y,int c,int l,int r,int rt){if(x<=l&&y>=r){add[rt]+=c;sum[rt]+=(LL)c*(r-l+1);return;}pushdown(rt,r-l+1);int m=(l+r)>>1;if(x<=m)update(x,y,c,lson);if(y>m)update(x,y,c,rson);pushup(rt);}LL query(int x,int y,int l,int r,int rt){if(x<=l&&y>=r)return sum[rt];pushdown(rt,r-l+1);int m=(l+r)>>1;LL ans=0;if(x<=m)ans+=query(x,y,lson);if(y>m)ans+=query(x,y,rson);return ans;}int main(){int n,q;scanf("%d%d",&n,&q);build(1,n,1);while(q--){char op[2];int a,b,c;scanf("%s",op);if(op[0]=='Q'){scanf("%d%d",&a,&b);printf("%lld\n",query(a,b,1,n,1));}else {scanf("%d%d%d",&a,&b,&c);update(a,b,c,1,n,1);}}}