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(线段树成段更新+区间求和) poj 3468

D - A Simple Problem with Integers
Time Limit:5000MS     Memory Limit:131072KB     64bit IO Format:%I64d & %I64u
Submit Status Practice POJ 3468

Description

You have N integers, A1A2, ... , AN. 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 A1A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C abc" means adding c to each of AaAa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q ab" means querying the sum of AaAa+1, ... , Ab.

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
#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<cstdlib>#include<algorithm>using namespace std;#define N 100010#define LL long long#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1LL add[N<<2],sum[N<<2];int n,m;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("%I64d",&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]+=(m-(m>>1))*add[rt];         sum[rt<<1|1]+=(m>>1)*add[rt];         add[rt]=0;      }}void update(int L,int R,int c,int l,int r,int rt){      if(L<=l&&r<=R)      {           add[rt]+=c;           sum[rt]+=(LL)(r-l+1)*c;           return ;      }      pushdown(rt,r-l+1);      int m=(l+r)>>1;      if(L<=m) update(L,R,c,lson);      if(m<R) update(L,R,c,rson);      pushup(rt);}LL query(int L,int R,int l,int r,int rt){      if(L<=l&&r<=R)      {            return sum[rt];      }      pushdown(rt,r-l+1);      int m=(l+r)>>1;      LL ret=0;      if(L<=m) ret+=query(L,R,lson);      if(m<R) ret+=query(L,R,rson);      return ret;}int main(){     scanf("%d%d",&n,&m);     build(1,n,1);     while(m--)     {           int a,b,c;           char s[5];           scanf("%s",s);           if(s[0]==‘Q‘)           {               scanf("%d%d",&a,&b);               printf("%I64d\n",query(a,b,1,n,1));           }           else           {                 scanf("%d%d%d",&a,&b,&c);                 update(a,b,c,1,n,1);           }     }     return 0;}

  

(线段树成段更新+区间求和) poj 3468