省选测试32

A 跑步

题目大意 ： 一个矩阵，求出从每个位置走到左上角可以获得的最大获利的和，动态修改，强制在线

• 首先容易求出每个点(i,j)到(1,1)的最大获利f[i][j]

• 每次只会改变一，所以f[i][j]如果要么改变一，要么不变，而且对于第i行改变的区域 \(l_i,r_i\) 具有单调性，也就是说一定有 \(l_{i-1}\leq l_i, r_{i-1}\leq r_i\)

• 有单调性就可以双指针了，给一个区间的f值同时加一，可以用差分，树状数组维护差分序列，这样就可以通过上面点的值和左面的之判断这个点的f值是否改变了

Code

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#include <cstdio>
#include <algorithm>

using namespace std;
const int N = 2005;

int read(int x = 0, int f = 1, char c = getchar()) {
    for (; c < '0' || c > '9'; c = getchar()) if (c == '-') f = -1;
    for (; c >='0' && c <='9'; c = getchar()) x = x * 10 + c - '0';
    return x * f;
}

int n, a[N][N], f[N][N], t[N][N];
long long ans;

void Add(int op, int x, int w) {
    for (; x <= n; x += x & -x) t[op][x] += w;
}

int Ask(int op, int x, int w = 0) {
    for (; x; x -= x & -x) w += t[op][x];
    return w;
}

bool Judge(int x, int y) {
    return Ask(x, y) == max(Ask(x-1, y), Ask(x, y-1)) + a[x][y];
}

int main() {
    n = read();
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            a[i][j] = read();
            f[i][j] = max(f[i-1][j], f[i][j-1]) + a[i][j];
            ans += f[i][j];
            Add(i, j, f[i][j] - f[i][j-1]);
        }
    }
    for (int i = 1; i <= n; ++i) t[i][n+1] = 1e9;
    printf("%lld\n", ans);
    for (int k = 1; k <= n; ++k) {
        char c; scanf(" %c", &c);
        int x = read(), y = read(), w = c == 'U' ? 1 : -1;
        int l = y, r = y; a[x][y] += w;
        for (int i = x; i <= n; ++i) {
            while (l <= n && Judge(i, l)) l++;
            Add(i, l, w); Add(i, r+1, -w); ans += w * (r - l + 1);
            while (r < n && !Judge(i, r+1)) r++, Add(i, r, w), Add(i, r+1, -w), ans += w;
        }
        printf("%lld\n", ans);
    }
    return 0;
}

B 算术

题目大意 ： 问一个大整数开k次方后是否为整数

• 随机个P，做k次剩余

Code

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#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

char c[1000005];
int len, k, M;

bool Prime(int x) {
    for (int i = 2; i * i <= x; ++i)
        if (x % i == 0) return 0;
    return 1;
}

int Pow(int a, int k, int ans = 1) {
    for (; k; k >>= 1, a = 1ll * a * a % M)
        if (k & 1) ans = 1ll * ans * a % M;
    return ans;
}

bool Judge() {
    for (int i = 1; i <= 20; ++i) {
        if (!Prime(M = i * k + 1)) continue;
        int s = 0;
        for (int j = 1; j <= len; ++j)
            s = (s * 10ll + c[j] - '0') % M;
        if (Pow(s, i) > 1) return 0;
    }
    return 1;
}

int main() {
    int T; scanf("%d", &T);
    while (T--) {
        scanf("%s%d", c + 1, &k);
        len = strlen(c + 1);
        puts(Judge() ? "Y" : "N");
    }
    return 0;
}

C 求和 (Unaccepted)

题目大意 ：

Code

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