树状数组 && 线段树

树状数组

支持单点修改

#include <cstdio>

using namespace std;

int n, m;
int a[], c[];

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

int sum(int x)
{
    int ans = ;
    while(x)
    {
        ans += c[x];
        x -= lowbit(x);
    }
    return ans;
}

void add(int x, int d)
{
    while(x <= n)
    {
        c[x] += d;
        x += lowbit(x);
    }
}

int main()
{
    int i, j, x, y, z;
    scanf("%d %d", &n, &m);
    for(i = ; i <= n; i++)
    {
        scanf("%d", &a[i]);
        add(i, a[i]);
    }
    for(i = ; i <= m; i++)
    {
        scanf("%d %d %d", &z, &x, &y);
        if(z == ) add(x, y);
        else printf("%d\n", sum(y) - sum(x - ));
    }
    return ;
}

支持区间修改

#include <cstdio>
#include <iostream>

using namespace std;

int n, m;
long long c0[], c1[], a[];

long long lowbit(int x) {return x & -x;}

long long sum(long long *c, int x)
{
    long long ans = ;
    while(x)
    {
        ans += c[x];
        x -= lowbit(x);
    }
    return ans;
}

void add(long long *c, int x, int d)
{
    while(x <= n)
    {
        c[x] += d;
        x += lowbit(x);
    }
}

int main()
{
    int i, j, x, y, z, k;
    long long ans;
    scanf("%d%d", &n, &m);
    for(i = ; i <= n; i++)
    {
        scanf("%d", &a[i]);
        add(c0, i, a[i]);
    }
    for(i = ; i <= m; i++)
    {
        scanf("%d%d%d", &z, &x, &y);
        if(z == )
        {
            scanf("%d", &k);
            add(c0, x, -k * (x - ));
            add(c1, x, k);
            add(c0, y + , k * y);
            add(c1, y + , -k);
        }
        else
        {
            ans = ;
            ans += sum(c0, y) + sum(c1, y) * y;
            ans -= sum(c0, x - ) + sum(c1, x - ) * (x - );
            printf("%lld\n", ans);
        }
    }
    return ;
}

线段树

支持区间修改

add[o]表示节点o的lazy标记，且节点o已经修改完

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;

#define root 1, 1, N
#define ls o << 1, l, m
#define rs o << 1 | 1, m + 1, r

int L, R;
long long add[], mul[], c[], P;

inline int read()
{
    int x = , f = ;
    char ch = getchar();
    while(!isdigit(ch))
    {
        if(ch == '-') f = -;
        ch = getchar();
    }
    while(isdigit(ch))
    {
        x = x *  + ch - '';
        ch = getchar();
    }
    return x * f;
}

inline void pushup(int o)
{
    c[o] = (c[o << ] + c[o <<  | ]) % P;
}

inline void build(int o, int l, int r)
{
    add[o] = ;
    mul[o] = ;
    if(l == r)
    {
        scanf("%lld", &c[o]);
        return;
    }
    int m = (l + r) >> ;
    build(ls);
    build(rs);
    pushup(o);
}

inline void pushdown(int o, int m)
{
    if(add[o] ==  && mul[o] == ) return;
    c[o << ] = (c[o << ] * mul[o] + add[o] * (m - (m >> ))) % P;
    c[o <<  | ] = (c[o <<  | ] * mul[o] + add[o] * (m >> )) % P;
    add[o << ] = (add[o << ] * mul[o] + add[o]) % P;
    add[o <<  | ] = (add[o <<  | ] * mul[o] + add[o]) % P;
    mul[o << ] = (mul[o << ] * mul[o]) % P;
    mul[o <<  | ] = (mul[o <<  | ] * mul[o]) % P;
    add[o] = ;
    mul[o] = ;
}

inline void update(int f, int d, int o, int l, int r)
{
    if(L <= l && r <= R)
    {
        if(f == )
        {
            add[o] = (add[o] + d) % P;
            c[o] = (c[o] + d * (r - l + )) % P;
        }
        else
        {
            mul[o] = (mul[o] * d) % P;
            add[o] = (add[o] * d) % P;
            c[o] = (c[o] * d) % P;
        }
        return;
    }
    pushdown(o, r - l + );
    int m = (l + r) >> ;
    if(L <= m) update(f, d, ls);
    if(m < R) update(f, d, rs);
    pushup(o);
}

inline long long query(int o, int l, int r)
{
    if(L <= l && r <= R) return c[o];
    pushdown(o, r - l + );
    int m = (l + r) >> ;
    long long ret = ;
    if(L <= m) ret += query(ls);
    if(m < R) ret += query(rs);
    return ret;
}

int main()
{
    int N, Q;
    N = read();
    P = read();
    build(root);
    Q = read();
    while(Q--)
    {
        int a, x, y, k;
        a = read();
        if(a ==  || a == )
        {
            x = read();
            y = read();
            k = read();
            L = x;
            R = y;
            update(a, k, root);
        }
        else
        {
            x = read();
            y = read();
            L = x;
            R = y;
            printf("%lld\n", query(root) % P);
        }
    }
    return ;
}

ps:有意思的是这个代码还是[AHOI2009]维护序列的题解