A Simple Problem with Integers（线段树 || 树状数组）

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 129659		Accepted: 40232
Case 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 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

Hint

The sums may exceed the range of 32-bit integers.

思路：模板题，网上都是详解，这里只贴个代码供自己翻阅。

树状数组区间修改区间查询，可以看这篇博客

线段树的学习，可以看这篇

树状数组：

#include<stdio.h>

#define For(a,b,c) for(int a = b; a <= c; a++)
#define ll long long

ll N, M;
ll c1[100005], c2[100005];

ll lowbit(ll x)
{
    return x&(-x);
}

void Add(ll *v, ll pos, ll num)
{
    for(; pos <= N; pos += lowbit(pos)) v[pos] += num;
}

ll Prefix_sum(ll *v, ll pos)
{
    ll ans = 0;
    for(; pos; pos-=lowbit(pos)) ans += v[pos];
    return ans;
}

ll sum(ll x)
{
    return x*Prefix_sum(c1,x) - Prefix_sum(c2,x);
}

ll Query(ll l, ll r)
{
    return sum(r) - sum(l-1);
}

int main()
{
    scanf("%lld %lld",&N,&M);
    ll last = 0, a;
    For(i,1,N)
    {
        scanf("%lld",&a);
        Add(c1,i,a-last);
        Add(c2,i,(i-1)*(a-last));
        last = a;
//        Add(c1,i+1,a);
//        Add(c2,i+1,-i*a);
    }
    char flag;
    ll l, r;
    while(M--)
    {
        scanf(" %c",&flag);
        if(flag == 'Q')
        {
            scanf("%lld%lld",&l,&r);
            printf("%lld\n",Query(l,r));
        }
        else
        {
            scanf("%lld%lld%lld",&l,&r,&a);
            Add(c1,l,a);
            Add(c1,r+1,-a);
            Add(c2,l,(l-1)*a);
            Add(c2,r+1,-r*a);
        }
    }
    return 0;
}

线段树：

#include<stdio.h>

#define For(a,b,c) for(int a = b; a <= c; a++)
#define ll long long

ll tree[400005], lazy[400005];

void Build(int l, int r, int rt)
{
    if(l == r)
    {
        scanf("%lld",&tree[rt]);
        return;
    }
    int mid = (l+r)>>1;
    Build(l, mid, rt<<1);
    Build(mid+1, r, rt<<1|1);
    tree[rt] = tree[rt<<1] + tree[rt<<1|1];
}

void pushdown(int rt, int len)
{
    if(lazy[rt])
    {
        lazy[rt<<1] += lazy[rt];
        lazy[rt<<1|1] += lazy[rt];

        tree[rt<<1] += lazy[rt]*(len-(len>>1));
        tree[rt<<1|1] += lazy[rt]*(len>>1);

        lazy[rt] = 0;
    }
}

void update(int l, int r, int ql, int qr, int rt, ll w)
{
    if(ql <= l && qr >= r)
    {
        tree[rt] += w*(r-l+1);
        lazy[rt] += w;
        return;
    }
    pushdown(rt,r-l+1);
    int mid = (l+r)>>1;
    if(ql <= mid) update(l,mid,ql,qr,rt<<1,w);
    if(qr > mid) update(mid+1,r,ql,qr,rt<<1|1,w);
    tree[rt] = tree[rt<<1] + tree[rt<<1|1];
}

ll Query(int l, int r, int ql, int qr, int rt)
{
//    printf("%d %d\n",l,r);
    if(ql <= l && r <= qr) return tree[rt];
    pushdown(rt,r-l+1);
    ll ans = 0;
    int mid = (l+r)>>1;
    if(ql <= mid) ans += Query(l,mid,ql,qr,rt<<1);
    if(qr > mid) ans += Query(mid+1,r,ql,qr,rt<<1|1);
    return ans;
}

int main()
{
    int N, Q;
    scanf("%d%d",&N,&Q);
    Build(1,N,1);
    char flag;
    int l, r;
    ll w;
    while(Q--)
    {
        scanf(" %c",&flag);
        if(flag == 'C')
        {
            scanf("%d%d%lld",&l,&r,&w);
            update(1,N,l,r,1,w);
        }
        else
        {
            scanf("%d%d",&l,&r);
            printf("%lld\n",Query(1,N,l,r,1));
        }
    }
    return 0;
}

A Simple Problem with Integers（线段树 || 树状数组）

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