A、完全平方数的尾巴

题目描述

示例1
输入
24
输出
true
备注:
1024 = 32 *32
24 = 1024 mod 1000

Solution

const int N = 1e3+7;
class Solution {
public:
    /**
     * 
     * @param x int整型 
     * @return bool布尔型
     */
    bool vis[N];
    bool solve(int x) {
        // write code here
        memset(vis,0,sizeof(vis));
        for(int i=0;i<N;++i)    vis[i*i%1000]=1;
        return vis[x];
    }
};

B、牛牛的字符串

#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(__vv__) (__vv__).begin(), (__vv__).end()
#define endl "\n"
#define pai pair<int, int>
#define mk(__x__,__y__) make_pair(__x__,__y__)
#define ms(__x__,__val__) memset(__x__, __val__, sizeof(__x__))
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void print(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }
const int dir[][2] = { {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;

const int N = 1e5 + 7;
int c[N][30];

void add(int p, int x, int k) {
    while (x <= 26) {
        c[p][x] += k;
        x += lowbit(x);
    }
}

int query(int p, int x) {
    int res = 0;
    while (x) {
        res += c[p][x];
        x -= lowbit(x);
    }
    return res;
}

class Solution {
public:
    /**
     *
     * @param s string字符串 s.size() <= 1e5
     * @param k int整型 k <= s.size()
     * @return int整型
     */
    int turn(string s, int k) {
        // write code here
        ms(c, 0);
        int n = s.size(), ans = 0;
        for (int i = 0; i < n; ++i) {
            int p = i % k;
            ans += query(p, s[i] - 'a');
            add(p, s[i] - 'a' + 1, 1);
        }
        return ans;
    }
};

C、下棋

二维前缀和，待补