POJ 3468 A Simple Problem with Integers

线段树的成段更新，区间求和基础题。

/*Accepted    4284K    1750MS    C++    1946B    2012-07-24 14:01:14*/

#include<cstdio>



#define lson l, m, rt << 1

#define rson m + 1, r, rt << 1 | 1

typedef long long LL;

const int MAXN = 100005;

LL add[MAXN << 2];

LL sum[MAXN << 2];



void PushUp( int rt )

{

    sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];

}



void PushDown( int l, int r, int rt)

{

    int ls = rt << 1, rs = rt << 1 | 1, m = (l + r) >> 1;

    if( add[rt])

    {

        add[ls] += add[rt];

        add[rs] += add[rt];

        sum[ls] += add[rt] * ( m - l + 1);

        sum[rs] += add[rt] * ( r - m);

        add[rt] = 0;

    }

}



void build( int l, int r, int rt)

{

    add[rt] = 0;

    if( l == r) {

        scanf( "%lld", &sum[rt]);

        return;

    }

    int m = (l + r) >> 1;

    build(lson);

    build(rson);

    PushUp(rt);

}



void update( int L, int R, LL c, int l, int r, int rt)

{

    int m = (l + r) >> 1;

    if( L <= l && r <= R) {

        add[rt] += c;

        sum[rt] += ( r - l + 1) * c;

        return;

    }

    PushDown( l, r, rt);

    if( L <= m) update(L, R, c, lson);

    if( R > m) update(L, R, c, rson);

    PushUp(rt);

}



LL query( int L, int R, int l, int r, int rt)

{

    int m = (l + r) >> 1;

    LL ret = 0;

    if( L <= l && r <= R) {

        return sum[rt];

    }

    PushDown(l, r, rt);

    if( L <= m) ret += query(L, R, lson);

    if( R > m) ret += query(L, R, rson);

    return ret;

}



int main()

{

    int n, q;

    while( scanf( "%d%d", &n, &q) == 2)

    {

        build( 1, n, 1);

        while( q --)

        {

            char s[2];

            int a, b;

            LL c;

            scanf( "%s", s);

            if( 'C' == s[0]){

                scanf( "%d%d%lld", &a, &b, &c);

                update( a, b, c, 1, n, 1);

            }

            if( 'Q' == s[0]){

                scanf( "%d%d", &a, &b);

                printf( "%lld\n", query( a, b, 1, n, 1));

            }

        }

    }

    return 0;

}