编程之美2015资格赛

标签：

A：处理下日期，容斥加减一下

B：DP，dp[l][r]表示区间回文子序列个数

C：模拟退火过了，然后还有个比较科学的方法，就是枚举B点，XY轴分开考虑，三分求解

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <string>
using namespace std;

int t, year;

char m1[2][15];
int m[2], d[2], y[2];

char sb[13][15] = {"", "January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November" , "December"};
map<string, int> to;

int main() {
    int cas = 0;
    for (int i = 1; i <= 12; i++)
        to[sb[i]] = i;
    scanf("%d", &t);
    while (t--) {
        scanf("%s%d,%d", m1[0], &d[0], &y[0]);
        scanf("%s%d,%d", m1[1], &d[1], &y[1]);
        m[0] = to[m1[0]]; m[1] = to[m1[1]];
        if (m[0] > 2 || (m[0] == 2 && d[0] > 29)) y[0]++;
        if (m[1] < 2 || (m[1] == 2 && d[1] < 29)) y[1]--;
        int ans = 0;
        ans += y[1] / 4 - (y[0] - 1) / 4;
        ans -= y[1] / 100 - (y[0] - 1) / 100;
        ans += y[1] / 400 - (y[0] - 1) / 400;
        printf("Case #%d: %d\n", ++cas, ans);
    }
    return 0;
}

#include <cstdio>
#include <cstring>

const int MOD = 100007;
const int N = 1005;
char str[N];
int t, dp[N][N];

int dfs(int l, int r) {
    if (dp[l][r] != -1) return dp[l][r];
    dp[l][r] = 0;
    if (l > r) return dp[l][r] = 0;
    if (l == r) return dp[l][r] = 1;
    if (str[l] == str[r]) dp[l][r] = (dp[l][r] + dfs(l + 1, r - 1) + 1) % MOD;
    dp[l][r] = ((dp[l][r] + dfs(l + 1, r) + dfs(l, r - 1) - dfs(l + 1, r - 1)) % MOD + MOD) % MOD;
    return dp[l][r];
}

int main() {
    int cas = 0;
    scanf("%d", &t);
    while (t--) {
        int ans = 0;
        scanf("%s", str);
        int n = strlen(str);
        memset(dp, -1, sizeof(dp));
        printf("%d\n", dfs(0, n - 1));
    }
    return 0;
}

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 105;

int t, n, m;
int an, bn;

struct Point {
    int x, y;
    Point() {}
    Point(int x, int y) {
        this->x = x;
        this->y = y;
    }
    void read() {
        scanf("%d%d", &x, &y);
    }
} a[N * 10], b[N];

const int INF = 0x3f3f3f3f;
const int d[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};
typedef long long ll;

ll cal(int x, int y) {
    ll ans = 0;
    for (int i = 0; i < an; i++) {
        int dx = x - a[i].x;
        int dy = y - a[i].y;
        ans += (ll)dx * dx;
        ans += (ll)dy * dy;
    }
    int Min = INF;
    for (int i = 0; i < bn; i++) {
        int dx = b[i].x - x;
        if (dx < 0) dx = -dx;
        int dy = b[i].y - y;
        if (dy < 0) dy = -dy;
        Min = min(Min, dy + dx);
    }
    return ans + Min;
}

int main() {
    int cas = 0;
    scanf("%d", &t);
    while (t--) {
        scanf("%d%d%d%d", &n, &m, &an, &bn);
        for (int i = 0; i < an; i++) a[i].read();
        for (int i = 0; i < bn; i++) b[i].read();
        int sx = 0, sy = 0;
        ll ans = (1LL<<63) - 1;
        int step = min(n, m);
        for (int ti = 0; ti < 1000; ti++) {
            for (int i = 0; i < 4; i++) {
                int x = sx + d[i][0] * step;
                int y = sy + d[i][1] * step;
                if (x < 0 || x > n || y < 0 || y > n) continue;
                ll tmp = cal(x, y);
                if (tmp < ans) {
                    sx = x;
                    sy = y;
                    ans = tmp;
                }
            }
            step = (step * 0.83);
            if (step == 0) step = 1;
        }
        printf("Case #%d: %lld\n", ++cas, ans);
    }
    return 0;
}

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long ll;
const int N = 105;

int ax[N * 10], ay[N * 10];

int t, n, m, an, bn;

int abss(int x) {
    if (x < 0) return -x;
    return x;
}

ll cal2(int u, int v, int *x) {
    ll ans = 0;
    ans += abss(u - v);
    for (int i = 0; i < an; i++) {
       int dx = u - x[i];
       ans += (ll)dx * dx;
    }
    return ans;
}

ll cal(int l, int r, int v, int *x) {
    while (r - l > 2) {
        int midl = (l * 2 + r) / 3;
        int midr = (l + 2 * r) / 3;
        if (cal2(midl, v, x) < cal2(midr, v, x)) r = midr;
        else l = midl;
    }
    ll ans = (1LL<<63) - 1;
    for (int i = l; i <= r; i++) ans = min(ans, cal2(i, v, x));
    return ans;
}

ll gao(int x, int y) {
    ll ans = 0;
    ans += min(cal(1, x, x, ax), cal(x, n, x, ax));
    ans += min(cal(1, y, y, ay), cal(y, m, y, ay));
    return ans;
}

int main() {
    int cas = 0;
    scanf("%d", &t);
    while (t--) {
        scanf("%d%d%d%d", &n, &m, &an, &bn);
        for (int i = 0; i < an; i++) scanf("%d%d", &ax[i], &ay[i]);
        ll ans = (1LL<<63) - 1;
        int x, y;
        for (int i = 0; i < bn; i++) {
            scanf("%d%d", &x, &y);
            ans = min(ans, gao(x, y));
        }
        printf("Case #%d: %lld\n", ++cas, ans);
    }
    return 0;
}

编程之美2015资格赛

标签：

原文地址：http://blog.csdn.net/accelerator_/article/details/45145735