炸鸡块君与FIFA22 线段树（牛客）

原题链接：登录—专业IT笔试面试备考平台_牛客网

 维护一个线段树的三种不同状态

 树的大小是点数的4倍，记得数组要开四倍

AC代码：

#include
using namespace std;
#define INF 0x3f3f3f3f
typedef pair PII;
const double pi = acos(-1.0);
#define rep(i, n) for (int i = 1; i <= (n); ++i)
#define rrep(i, n) for (int i = n; i >= (1); --i)
typedef long long ll;
#define sqar(x) ((x)*(x))

const int N = 2e5 + 10;
char s[N];

struct
{
    int l, r;
    int c[3]; //l到r区间以i为初始值（余3的余数）区间的贡献
}tree[N << 2];

void update(int num)
{
    for(int i = 0; i < 3; i++) tree[num].c[i] = tree[num << 1].c[i] + tree[num << 1 | 1].c[(i + tree[num << 1].c[i]) % 3];
}

void build_tree(int num, int l, int r)
{
    tree[num].l = l, tree[num].r = r;
    if(l == r)
    {
        if(s[l] == 'W')
        {
             tree[num].c[0] = 1;
             tree[num].c[1] = 1;
             tree[num].c[2] = 1;
        }
        else if(s[l] == 'L')
        {
            tree[num].c[0] = 0;
            tree[num].c[1] = -1;
            tree[num].c[2] = -1;
        }
        else{
            tree[num].c[0] = 0;
            tree[num].c[1] = 0;
            tree[num].c[2] = 0;
        }
    }
    else{
        int mid = (l + r) >> 1;
        build_tree(num << 1, l, mid);
        build_tree(num << 1 | 1, mid + 1, r);
        update(num);
    }
}

int query(int num, int l, int r, int val)
{
    if(tree[num].l == l && tree[num].r == r) return tree[num].c[val];
    int mid = (tree[num].l + tree[num].r) >> 1;
    int sum = 0;
    if(l <= mid)
    {
        sum += query(num << 1, l, min(mid, r), val);
    }
    if(r > mid)
    {
        sum += query(num << 1 | 1, max(mid + 1, l), r, (val + sum) % 3);
    }
    return sum;
}

int main()
{
    int n, q;
    scanf("%d %d", &n, &q);
    scanf("%s", s + 1);
    build_tree(1, 1, n);
    rep(i, q)
    {
        int l, r, s;
        scanf("%d %d %d", &l, &r, &s);
        ll res = s + query(1, l, r, s % 3);
        printf("%lldn", res);
    }
    return 0;
}

 两个小时前学的线段树，看了大佬的题解之后自己写了一遍就过了，我好像变聪明了哈哈