POJ A Simple Problem with Integers 线段树 lazy-target 区间跟新

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 105742 Accepted: 33031Case 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

思路：线段树区间跟新 + lazy-target标记

（注意数据范围，long long）

代码：

#include<iostream>#include<cstring>#include<string>#include<cmath>#include<algorithm>#include<cstdio>using namespace std;#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1const int maxn=100005;long long sum[maxn<<2];long long add[maxn<<2];void pushup(int rt) {    sum[rt]=sum[rt<<1]+sum[rt<<1|1];}void pushdown(int rt, int len) {    if(add[rt]) {        add[rt<<1]+=add[rt];        add[rt<<1|1]+=add[rt];        sum[rt<<1]+=(len-(len>>1))*add[rt];        sum[rt<<1|1]+=(len>>1)*add[rt];        add[rt]=0;    }}void build(int l, int r, int rt) {    add[rt]=0;    if(l==r) {        scanf("%lld",&sum[rt]);        return;    }    int mid=(l+r)>>1;    build(lson);    build(rson);    pushup(rt);}void update(int L, int R, int val, int l, int r, int rt) {    if(L<=l&&r<=R) {        sum[rt]+=(r-l+1)*val;        add[rt]+=val;        return;    }    pushdown(rt,r-l+1);    int mid=(l+r)>>1;    if(L<=mid) update(L,R,val,lson);    if(R>mid) update(L,R,val,rson);    pushup(rt);}long long 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 mid=(l+r)>>1;    long long cnt=0;    if(L<=mid) cnt+=query(L,R,lson);    if(R>mid) cnt+=query(L,R,rson);    return cnt;}int main() {    int n,q;    while(~scanf("%d%d",&n,&q)) {        build(1,n,1);        char s[10];        for(int i=1;i<=q;i++) {            scanf("%s",s);            if(s[0]=='Q') {                int L,R;long long result;scanf("%d%d",&L,&R);                result=query(L,R,1,n,1);                printf("%lld\n",result);            } else if(s[0]=='C') {                int L,R,val;scanf("%d%d%d",&L,&R,&val);                update(L,R,val,1,n,1);            }        }    }    return 0;}