C - A Simple Problem with Integers POJ 3468（线段树+延迟标记）

C - 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, A1, A2, ... , 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 A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C abc" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q ab" means querying the sum of Aa, Aa+1, ... , Ab.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4

Sample Output

4
55
9

15

//线段树 的区间更新，用好延迟标记

//#include<bits/stdc++.h>#include<cstdio>#include<algorithm>using namespace std;const int maxn=500011;const int inf=999999999;#define lson (rt<<1),L,M#define rson (rt<<1|1),M+1,R#define M ((L+R)>>1)#define For(i,n) for(int i=0;i<(n);i++)template<class T>inline T read(T&x){    char c;    while((c=getchar())<=32);    bool ok=false;    if(c=='-')ok=true,c=getchar();    for(x=0; c>32; c=getchar())        x=x*10+c-'0';    if(ok)x=-x;    return x;}template<class T> inline void read_(T&x,T&y){    read(x);    read(y);}template<class T> inline void write(T x){    if(x<0)putchar('-'),x=-x;    if(x<10)putchar(x+'0');    else write(x/10),putchar(x%10+'0');}template<class T>inline void writeln(T x){    write(x);    putchar('\n');}//-------IO template------typedef __int64 LL;struct node{    LL sum;    LL inv;    node(){sum=0;inv=0;}}p[maxn<<2];void build(LL rt,LL L,LL R){    if(L==R)    {        scanf("%I64d",&p[rt].sum);//read(p[rt].sum);        return;    }    build(lson);    build(rson);    p[rt].sum=p[rt<<1].sum+p[rt<<1|1].sum;}void update(LL rt,LL L,LL R,LL x,LL y,LL add){    if(L==x&&R==y)    {        p[rt].inv+=add;        p[rt].sum+=(LL)add*(R-L+1);        return;    }    if(p[rt].inv)///p[rt].inv可能收负数，之前写成了p[rt].inv>0 导致WA很多次    {        p[rt<<1].inv+=p[rt].inv;        p[rt<<1|1].inv+=p[rt].inv;        p[rt<<1].sum+=(LL)p[rt].inv*(M-L+1);        p[rt<<1|1].sum+=(LL)p[rt].inv*(R-M);        p[rt].inv=0;    }    if(y<=M)        update(lson,x,y,add);    else if(x>M)        update(rson,x,y,add);    else{        update(lson,x,M,add);        update(rson,M+1,y,add);    }    p[rt].sum=p[rt<<1].sum+p[rt<<1|1].sum;}LL query(LL rt,LL L,LL R,LL x,LL y){    if(x==L&&y==R)    {        return p[rt].sum;//+p[rt].inv*(R-L+1);    }     if(p[rt].inv)     {        p[rt<<1].inv+=p[rt].inv;        p[rt<<1|1].inv+=p[rt].inv;        p[rt<<1].sum+=(LL)p[rt].inv*(M-L+1);        p[rt<<1|1].sum+=(LL)p[rt].inv*(R-M);        p[rt].inv=0;     }    if(y<=M)        return query(lson,x,y);    else if(x>M)        return query(rson,x,y);    else return query(lson,x,M)+query(rson,M+1,y);}int main(){    //#ifndef ONLINE_JUDGE   // freopen("in.txt","r",stdin);   // #endif // ONLINE_JUDGE    LL m,n,i,j,k,t;    while(~scanf("%I64d%I64d",&n,&m))    {        build(1,1,n);        char op[2];        while(m--)        {            LL  x,y;            scanf("%s",op);scanf("%I64d%I64d",&x,&y);            if(op[0]=='Q')            {                printf("%I64d\n",query(1,1,n,x,y));            }            else{                LL z;                scanf("%I64d",&z);                update(1,1,n,x,y,z);            }        }    }    return 0;}