BestCoder #18

最新推荐文章于 2017-08-17 02:06:10 发布

转载最新推荐文章于 2017-08-17 02:06:10 发布 · 50 阅读

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原文链接：http://www.cnblogs.com/rolight/p/4102498.html

本文解析了四道编程题目，包括素数问题、求导极值、数位DP及线段树应用。提供了完整的代码实现，并针对每道题目进行了详细的说明。

A 傻缺题，直接先把素数搞出来然后枚举一下就好了。

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

using namespace std;
const int maxn = 1e4 + 10;
bool vis[maxn];
vector<int> pnum;

void init() {
    vis[1] = true;
    int maxv = 10000;
    for(int i = 2; i <= maxv; i++) if(!vis[i]) {
        pnum.push_back(i);
        for(int j = i + i; j <= maxv; j += i) vis[j] = true;
    }
}

void gao(int num) {
    int ans = 0;
    int msize = pnum.size();
    for(int i = 0; i < msize; i++) {
        for(int j = i; j < msize && pnum[i] + pnum[j] < num; j++) {
            int tt = num - pnum[i] - pnum[j];
            if(!vis[tt] && tt >= pnum[j]) ans++;
            if(tt < pnum[j]) break;
        }
    }
    printf("%d\n", ans);
}

int main() {
    init();
    int n;
    while(scanf("%d", &n) != EOF) {
        gao(n);
    }
    return 0;
}

　B 直接求导之后找极值点就好了，注意特判a,b等于0的情况。

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

using namespace std;

const double eps = 1e-9;

int getx2p(double a, double b, double c, double &x1, double &x2) {
    double delta = b * b - 4 * a * c;
    if(delta < 0) return 0;
    x1 = -b + sqrt(delta); x2 = -b - sqrt(delta);
    x1 /= 2 * a; x2 /= 2 * a;
    if(delta < eps) return 1;
    else return 2;
}

bool inrange(double n, double L, double R) {
    L -= eps; R += eps;
    return (n >= L && n <= R);
}

double f(double x, double a, double b, double c, double d) {
    return a * x * x * x + b * x * x + c * x + d;
}

int main() {
    double a, b, c, d, L, R;
    while(scanf("%lf%lf%lf%lf%lf%lf", &a, &b, &c, &d, &L, &R) != EOF) {
        double p0, p1;
        int ret = getx2p(3 * a, 2 * b, c, p0, p1);
        double ans = -1e30;
		if(a == 0 && b == 0) {
            ans = max(abs(f(L, a, b, c, d)), abs(f(R, a, b, c, d)));
		}
		else if(a == 0) {
            ans = max(abs(f(L, a, b, c, d)), abs(f(R, a, b, c, d)));
			double mid = -c / (2 * b);
			if(inrange(mid, L, R)) {
				ans = max(ans, f(mid, a, b, c, d));
			}
		}
		else if(ret == 0 || (!inrange(p0, L, R) && !inrange(p1, L, R))) {
            ans = max(abs(f(L, a, b, c, d)), abs(f(R, a, b, c, d)));
        }
        else {
            ans = max(abs(f(L, a, b, c, d)), abs(f(R, a, b, c, d)));
            if(inrange(p0, L , R)) ans = max(ans, abs(f(p0, a, b, c, d)));
            if(inrange(p1, L , R)) ans = max(ans, abs(f(p1, a, b, c, d)));
        }
        printf("%.2f\n", ans);
    }
    return 0;
}

　C 数位DP，思路和HDU 4507很像，也可以用排列组合直接搞。。

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

using namespace std;

typedef long long LL;
const int maxn = 1e3 + 100;
const LL mod = 1e9 + 7;
char buf[maxn];
int lim[maxn], n, len;
LL p2[maxn];
bool vis[maxn][maxn];

struct EE {
    LL cnt, sum;
    EE(LL cnt = 0, LL sum = 0): cnt(cnt), sum(sum) {}
};

EE f[maxn][maxn];

EE dfs(int now, int bitcnt, bool bound) {
    if(now == 0) return EE(bitcnt == 0, 0);
    if(vis[now][bitcnt] && !bound) return f[now][bitcnt];
    int m = bound ? lim[now - 1] : 1;
    EE ret(0, 0);
    for(int i = 0; i <= m; i++) {
        EE nret = dfs(now - 1, bitcnt - i, bound && i == m);
        LL ff = p2[now - 1] * i % mod;
        ret.cnt = (ret.cnt + nret.cnt) % mod;
        ret.sum = ((nret.cnt * ff % mod + nret.sum) % mod + ret.sum) % mod;
    }
    if(!bound) {
        vis[now][bitcnt] = true;
        f[now][bitcnt] = ret;
    }
    return ret;
}

void gao() {
    len = strlen(buf);
    bool flag = false;
    for(int i = 0, j = len - 1; i < len; i++, j--) {
        lim[j] = buf[i] - '0';
    }
    int pos;
    for(pos = 0; lim[pos] == 0; pos++) lim[pos] = 1;
    lim[pos] = 0;
    if(pos == len - 1) len--;
    EE ret = dfs(len, n, true);
    cout << ret.sum  << endl;
}

void init() {
    p2[0] = 1;
    for(int i = 1; i <= 1000; i++) {
        p2[i] = (p2[i - 1] * 2) % mod;
    }
}

int main() {
    init();
    while(scanf("%d%s", &n, buf) != EOF) {
        gao();
    }
    return 0;
}

　D 线段树

先把所有的点按照y排序，然后在x轴上一个一个插入然后统计即可。

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

using namespace std;
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
const int maxn = 3e4 + 100;
const int inf = 1e9 + 100;

struct Node {
    int val[12];
    Node() {
        for(int i = 1; i <= 11; i++) val[i] = inf;
    }
    void Megre(int val1[12]) {
        int tar[22];
		for(int i = 1; i <= 10; i++) {
			tar[i] = val[i]; tar[i + 10] = val1[i];
		}
		sort(tar + 1, tar + 21);
        for(int i = 1; i <= 10; i++) val[i] = tar[i];
		val[11] = inf;
    }
	void Megre(int k) {
		val[11] = k;
		sort(val + 1, val + 12);
		val[11] = inf;
	}
	void print() {
		for(int i = 1; i <= 10; i++)  {
			printf("%d ", val[i]);
		}
		puts("");
	}
};

Node minv[maxn << 4];

void pushup(int rt) {
    minv[rt] = minv[rt << 1];
    minv[rt].Megre(minv[rt << 1 | 1].val);
}

void build(int rt, int l, int r) {
    if(l == r) minv[rt] = Node();
    else {
        int mid = (l + r) >> 1;
        build(lson); build(rson);
        pushup(rt);
    }
}

void update(int rt, int l, int r, int pos, int x) {
    if(l == r) minv[rt].Megre(x);
    else {
        int mid = (l + r) >> 1;
        if(pos <= mid) update(lson, pos, x);
        else update(rson, pos, x);
        pushup(rt);
    }
}

Node query(int rt, int l, int r, int ql, int qr) {
    if(ql <= l && qr >= r) return minv[rt];
    else {
        int mid = (l + r) >> 1;
        Node ret = Node();
        if(ql <= mid) ret.Megre(query(lson, ql, qr).val);
        if(qr > mid) ret.Megre(query(rson, ql, qr).val);
		return ret;
    }
}

struct Point {
    int x, y, h, k;
    int id;
    bool isq;
    Point(int x, int y, int h, bool isq, int k = 0, int id = 0):
        x(x), y(y), h(h), isq(isq), k(k), id(id) {}
    bool operator < (const Point &p) const {
        if(y == p.y) return x < p.x;
        return y < p.y;
    }
};

int n, m, ans[maxn];
vector<int> vnum;
vector<Point> pp;

int getid(int val) {
    int ret = lower_bound(vnum.begin(), vnum.end(), val) - vnum.begin() + 1;
	return ret;
}

int main() {
    while(scanf("%d%d", &n, &m) != EOF) {
        vnum.clear(); pp.clear();
        for(int i = 1; i <= n; i++) {
            int x, y, h; scanf("%d%d%d",&x, &y, &h);
            pp.push_back(Point(x, y, h, false));
            vnum.push_back(x);
        }
        for(int i = 1; i <= m; i++) {
            int x, y, k;
            scanf("%d%d%d", &x, &y, &k);
            pp.push_back(Point(x, y, 0, true, k, i));
            vnum.push_back(x);
        }
        sort(vnum.begin(), vnum.end());
        sort(pp.begin(), pp.end());
        vnum.erase(unique(vnum.begin(), vnum.end()), vnum.end());
        int mm = pp.size(), nn = vnum.size();
        build(1, 1, nn);
        for(int i = 0; i < mm; i++) {
            if(pp[i].isq == false) {
                update(1, 1, nn, getid(pp[i].x), pp[i].h);
            }
            else {
                Node ret = query(1, 1, nn, 1, getid(pp[i].x));
                ans[pp[i].id] = ret.val[pp[i].k];
            }
        }
        for(int i = 1; i <= m; i++) {
            if(ans[i] == inf) ans[i] = -1;
            printf("%d\n", ans[i]);
        }
    }
    return 0;
}

转载于:https://www.cnblogs.com/rolight/p/4102498.html