poj 3468A Simple Problem with Integers

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.

这题是线段树成段更新，看别人代码想了很久。

思路：开一个结构体，含有左右边界l,r，这段区间线段总和sum，更新标志ans（整段区间内每个数要加的数值）。

每次一个区间更新（即加一个数value)的时候，从第一个线线段开始向下判断，如果更新的区间刚好是这条线段的区间，那么直接加上更新的数值value并返回，否则整段区间的sum值变为sum+(b-a+1)*value，再在子节点中找直到找到区间大小加好符合的时候，ans=ans+value,返回。

每一次询问，从第一个线段开始向下，如果区间刚好符合，那么返回区间的sum否则把这条线段的更新标志往子节点传，同时这条线段的更新标志变为0.这里我采用的是每一次更新就把每根点段的总和都保存在sum中，这样询问的时候就不用麻烦的加上b[i].ans*(b[i].r-b[i].l+1).

#include<stdio.h>#include<string.h>#define maxn 100005#define ll long longchar str[10];ll a[maxn];struct node{ll l,r,sum,ans;}b[4*maxn];void build(ll l,ll r,ll i){    ll mid;    b[i].l=l;    b[i].r=r;    b[i].ans=0;    if(b[i].l==b[i].r)    {       b[i].sum=a[l];       return;    }    mid=(l+r)/2;    build(l,mid,2*i);    build(mid+1,r,i*2+1);    b[i].sum=b[i*2].sum+b[i*2+1].sum;}void pushdown(int i){    if(b[i].ans){        b[i*2].ans+=b[i].ans;        b[i*2+1].ans+=b[i].ans;        b[i*2].sum+=b[i].ans*(b[i*2].r-b[i*2].l+1);        b[i*2+1].sum+=b[i].ans*(b[i*2+1].r-b[i*2+1].l+1);        b[i].ans=0;    }}void add(ll l,ll r,ll value,ll i){ll mid;if(b[i].l==l && b[i].r==r){   b[i].ans=b[i].ans+value;   b[i].sum+=(b[i].r-b[i].l+1)*value;   return;}pushdown(i);b[i].sum+=(r-l+1)*value; //这一句写了，下面第二句就不用写了，是同一个意思mid=(b[i].l+b[i].r)/2;if(l>mid)add(l,r,value,i*2+1);else if(r<=mid)add(l,r,value,i*2);else{   add(l,mid,value,i*2);   add(mid+1,r,value,i*2+1);}//b[i].sum=b[i*2].sum+b[i*2+1].sum;  ---2}ll question(ll l,ll r,ll i){ll mid;if(b[i].l==l && b[i].r==r){   return b[i].sum;}pushdown(i);mid=(b[i].l+b[i].r)/2;if(l>mid)return question(l,r,i*2+1);else if(r<=mid)return question(l,r,i*2);else if(l<=mid && r>mid)return question(l,mid,i*2)+question(mid+1,r,i*2+1);}int main(){ll n,m,c,d,e,i,j;while(scanf("%lld%lld",&n,&m)!=EOF){for(i=1;i<=n;i++)scanf("%lld",&a[i]);build(1,n,1);while(m--){scanf("%s",str);if(str[0]=='Q'){scanf("%lld%lld",&c,&d);printf("%lld\n",question(c,d,1));}else if(str[0]=='C'){scanf("%lld%lld%lld",&c,&d,&e);add(c,d,e,1);}}}return 0;}