学步园

需求：区间更新，区间求和。

#include <algorithm>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>

#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <string>
#include <bitset>
#include <memory>
#include <complex>
#include <numeric>

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <time.h>
#include <ctype.h>
#include <locale.h>

using namespace std;

#pragma pack(4)

const double  eps = 1e-8;
const double   pi = acos(-1.0);
const int     inf = 0x7f7f7f7f;
const __int64 INF = 0x7f7f7f7f7f7f7f7fll;

#define loop(a,n)                            \
    for(int i=0;n>i;i++)                     \
        cout<<a[i]<<(i!=n-1?' ':'\n')
#define loop2(a,n,m)                         \
    for(int i=0;n>i;i++)                     \
        for(int j=0;m>j;j++)                 \
            cout<<a[i][j]<<(j!=m-1?' ':'\n')

#define   at(a,i) ((a)&(1<<(i)))
#define   nt(a,i) ((a)^(1<<(i)))
#define set1(a,i) ((a)|(1<<(i)))
#define set0(a,i) ((a)&(~(1<<(i))))

#define gret(a,b) (((a)-(b))>eps)
#define less(a,b) (((b)-(a))>eps)
#define greq(a,b) (((a)-(b))>-eps)
#define leeq(a,b) (((b)-(a))>-eps)
#define equl(a,b) (fabs((a)-(b))<eps)

#define lmax(a,b) ((a)>(b)?(a):(b))
#define lmin(a,b) ((a)<(b)?(a):(b))
#define fmax(a,b) (gret(a,b)?(a):(b))
#define fmin(a,b) (less(a,b)?(a):(b))

const int MAXV = 100002;

#define lc(i) ((i)<<1)
#define rc(i) ((i)<<1|1)

char o;
__int64 n,m,u,v,w,sum[MAXV<<2],lazy[MAXV<<2];

void down(__int64 i,__int64 n)
{
    lazy[lc(i)]+=lazy[i];
    lazy[rc(i)]+=lazy[i];
    sum[lc(i)]+=(lazy[i]*(n-(n>>1)));
    sum[rc(i)]+=(lazy[i]*(n>>1));
    lazy[i]=0;
}
void change(__int64 x,__int64 y,__int64 w,__int64 l=1,__int64 r=n,__int64 i=1)
{
    if(x<=l&&r<=y)
    {
        lazy[i]+=w;
        sum[i]+=w*(r-l+1);
    }
    else
    {
        if(lazy[i]) down(i,r-l+1);

        if(x<=(l+r)/2) change(x,y,w,l,(l+r)/2,lc(i));
        if((l+r)/2<y) change(x,y,w,(l+r)/2+1,r,rc(i));

        sum[i]=sum[lc(i)]+sum[rc(i)];
    }
}
__int64 query(__int64 x,__int64 y,__int64 l=1,__int64 r=n,__int64 i=1)
{
    if(x<=l&&r<=y) return sum[i];
    else
    {
        if(lazy[i]) down(i,r-l+1);

        __int64 cnt=0;
        if(x<=(l+r)/2) cnt+=query(x,y,l,(l+r)/2,lc(i));
        if((l+r)/2<y) cnt+=query(x,y,(l+r)/2+1,r,rc(i));

        return cnt;
    }
}

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("A Simple Problem with Integers.txt","r",stdin);
    #endif

    while(scanf("%I64d %I64d",&n,&m)!=EOF)
    {
        memset(sum,0,sizeof(sum));
        memset(lazy,0,sizeof(lazy));

        for(__int64 i=1;n>=i;i++)
        {
            scanf("%I64d",&w);
            change(i,i,w);
        }
        for(__int64 i=1;m>=i;i++)
        {
            getchar();
            scanf("%c",&o);
            if(o=='Q')
            {
                scanf("%I64d %I64d",&u,&v);
                printf("%I64d\n",query(u,v));
            }
            else if(o=='C')
            {
                scanf("%I64d %I64d %I64d",&u,&v,&w);
                change(u,v,w);
            }
        }
    }
    return 0;
}