poj 3468 A Simple Problem with Integers（线段树）

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A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 61636 Accepted: 18840Case 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

Hint

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

题解及代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <set>#include <map>#include <queue>#include <string>#define maxn 100010#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define ALL %I64dusing namespace std;typedef __int64  ll;struct segment{    int l,r;    ll value;    ll nv;} son[maxn<<2];void PushUp(int rt){    son[rt].value=son[rt<<1].value+son[rt<<1|1].value;}void Build(int l,int r,int rt){    son[rt].l=l;    son[rt].r=r;    son[rt].nv=0;    if(l==r)    {        scanf("%I64d",&son[rt].value);        return;    }    int m=(l+r)/2;    Build(lson);    Build(rson);    PushUp(rt);}void Update_n(ll w,int l,int r,int rt){    if(son[rt].l==l&&son[rt].r==r)    {        son[rt].value+=w*(r-l+1);        son[rt].nv+=w;        return;    }    if(son[rt].nv)    {        son[rt<<1].nv+=son[rt].nv;        son[rt<<1|1].nv+=son[rt].nv;        son[rt<<1].value+=(son[rt<<1].r-son[rt<<1].l+1)*son[rt].nv;        son[rt<<1|1].value+=(son[rt<<1|1].r-son[rt<<1|1].l+1)*son[rt].nv;        son[rt].nv=0;    }    int m=(son[rt].l+son[rt].r)/2;    if(r<=m)        Update_n(w,l,r,rt<<1);    else if(l>m)        Update_n(w,l,r,rt<<1|1);    else    {        Update_n(w,lson);        Update_n(w,rson);    }    PushUp(rt);}ll  Query(int l,int r,int rt){    if(son[rt].l==l&&son[rt].r==r)    {        return son[rt].value;    }    if(son[rt].nv)    {        son[rt<<1].nv+=son[rt].nv;        son[rt<<1|1].nv+=son[rt].nv;        son[rt<<1].value+=(son[rt<<1].r-son[rt<<1].l+1)*son[rt].nv;        son[rt<<1|1].value+=(son[rt<<1|1].r-son[rt<<1|1].l+1)*son[rt].nv;        son[rt].nv=0;    }    ll ret=0;    int m=(son[rt].l+son[rt].r)/2;    if(r<=m)        ret=Query(l,r,rt<<1);    else if(l>m)        ret=Query(l,r,rt<<1|1);    else    {        ret=Query(lson);        ret+=Query(rson);    }    //PushUp(rt);    return ret;}int main(){    int n,m,l,r;    ll w;    char s[4];    while(scanf("%d%d",&n,&m)!=EOF)    {        Build(1,n,1);        for(int i=1; i<=m; i++)        {            scanf("%s",s);            if(s[0]=='C')            {                scanf("%d%d%I64d",&l,&r,&w);                Update_n(w,l,r,1);            }            else            {                scanf("%d%d",&l,&r);                printf("%I64d\n",Query(l,r,1));            }        }    }    return 0;}/*作为初学者，写起程序也总是会出现一些小毛病，都是对于线段树的理解不深造成的。说一下这道题目需要注意的地方，那就是无论是每个节点的值还是增量，都要使用long long，否则会错掉。经过这段时间的学习，弄懂了线段树的好多地方，现在感觉做题也没那么费力了，都是简单题目，对于初学者来说还是比较好的，欢迎大家一起讨论学习。*/

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