poj 3468 Simple Problem with Integers(区间更新+懒惰标记)

 Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 109294 Accepted: 34031Case Time Limit: 2000MS

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

Hint

The sums may exceed the range of 32-bit integers.

tips:说下两个坑点吧.  

1.懒惰标记时应该是追加标记，而不是覆盖。

2.lazy的类型也应该是ll类型的，因为在pushdown的累加过程中可能超int(因为这个自己wa无数)

#include<iostream>#include<cstring>#include<cstdio>using namespace std;const int maxn=110000;#define ll long longint n,q;ll sum[maxn<<2];ll lazy[maxn<<2];void pushup(int rt){sum[rt]=sum[rt<<1]+sum[rt<<1|1];}void build(int rt,int l,int r){if(l==r){scanf("%lld",&sum[rt]);return;}int m=(l+r)>>1;build(rt<<1,l,m);build(rt<<1|1,m+1,r);pushup(rt);}void pushdown(int rt,int l,int r){if(lazy[rt]){lazy[rt<<1]+=lazy[rt];lazy[rt<<1|1]+=lazy[rt];int m=(l+r)>>1;sum[rt<<1]+=lazy[rt]*(m-l+1);sum[rt<<1|1]+=lazy[rt]*(r-m);lazy[rt]=0;}}void update(int x,int y,int z,int rt,int l,int r){if(x<=l&&y>=r){lazy[rt]+=z;sum[rt]+=(ll)z*(r-l+1);return;}pushdown(rt,l,r);int m=(l+r)>>1;if(x<=m)update(x,y,z,rt<<1,l,m);if(y>m)update(x,y,z,rt<<1|1,m+1,r);pushup(rt);}ll query(int x,int y,int rt,int l,int r){if(x<=l&&y>=r){return sum[rt];}pushdown(rt,l,r);ll ret=0;int m=(l+r)>>1;if(x<=m)ret+=query(x,y,rt<<1,l,m);if(y>m)ret+=query(x,y,rt<<1|1,m+1,r);return ret;}int main(){while(~scanf("%d %d",&n,&q)){memset(sum,0,sizeof(sum));memset(lazy,0,sizeof(lazy));//scanf("%d %d",&n,&q);build(1,1,n);char s[5];while(q--){scanf("%s",s);if(s[0]=='Q'){int x,y;scanf("%d %d",&x,&y); printf("%lld\n",query(x,y,1,1,n));}else{int x,y,z;scanf("%d %d %d",&x,&y,&z);update(x,y,z,1,1,n);}}}return 0; }