2017-09-29校训练题题解

t1

傻逼dfs，大暴搜，不贴代码。

t2

ans= 黑点数 − 边数，二维前缀和。

#include<set>#include<queue>#include<cstdio>#include<vector>#include<cstring>#include<iostream>#include<algorithm>using namespace std;inline int read() {    int x = 0, flag = 1; char ch = getchar();    while (ch > '9' || ch < '0') { if (ch == '-') flag = -1; ch = getchar(); }    while (ch <= '9' && ch >= '0') { x = x * 10 + ch - '0'; ch = getchar(); }    return x * flag;}inline bool get() {    char ch = getchar();    while (ch != '1' && ch != '0') ch = getchar();    return ch == '1';}#define rep(ii, aa, bb) for (int ii = aa; ii <=  bb; ii++)#define ll long long#define N 2001bool a[N][N];int sum[N][N], upe[N][N], lfe[N][N];int main() {    int n, m, q; scanf("%d%d%d", &n, &m, &q);    rep(i, 1, n) rep(j, 1, m) a[i][j] = get();    rep(i, 1, n) rep(j, 1, m) {        upe[i][j] = upe[i - 1][j] + (a[i][j] & a[i][j - 1]);        lfe[i][j] = lfe[i][j - 1] + (a[i][j] & a[i - 1][j]);        sum[i][j] = a[i][j] - (a[i][j] & a[i][j - 1]) - (a[i][j] & a[i - 1][j]) + sum[i][j - 1] + sum[i - 1][j] - sum[i - 1][j - 1];    }    while (q--) {        int x1 = read(), y1 = read(), x2 = read(), y2 = read();        int ans = 0;        ans += sum[x2][y2] - sum[x2][y1 - 1] - sum[x1 - 1][y2] + sum[x1 - 1][y1 - 1];        ans -= upe[x1 - 1][y1] - upe[x2][y1];        ans -= lfe[x1][y1 - 1] - lfe[x1][y2];        printf("%d\n", ans);    }    return 0;}

t3

卡空间的逆序对，玄学颓公式。

#include<set>#include<queue>#include<cstdio>#include<vector>#include<cstring>#include<iostream>#include<algorithm>using namespace std;inline int read() {    int x = 0, flag = 1; char ch = getchar();    while (ch > '9' || ch < '0') { if (ch == '-') flag = -1; ch = getchar(); }    while (ch <= '9' && ch >= '0') { x = x * 10 + ch - '0'; ch = getchar(); }    return x * flag;}#define rep(ii, aa, bb) for (int ii = aa; ii <=  bb; ii++)#define ll long long#define N 100001int n, x, a, mod, cnt = 0;struct period {    int base, cnt;    ll operator * (const period &t) {        int st = (base - t.base + a) / a;        int tn = t.cnt - st + 1;        if (tn >= cnt) return (ll)cnt * ((st << 1) + cnt - 1) >> 1;        return ((ll)tn * (st + t.cnt) >> 1) + (ll)t.cnt * (cnt - tn);    }} p[N];int main() {    scanf("%d%d%d%d", &n, &x, &a, &mod);    while (true) {        int tmp = (mod - x + a - 1) / a;        if (tmp >= n) {            p[cnt].base = x, p[cnt].cnt = n;            break;        }        p[cnt].base = x, p[cnt++].cnt = tmp;        x += tmp*a - mod;        n -= tmp;    }    ll ans = 0;    rep(i, 0, cnt - 1) rep(j, i + 1, cnt) ans += p[i] * p[j];    printf("%lld\n", ans);    return 0;}