Codeforces Round #296 (Div. 2) A B C D

标签：

A：模拟辗转相除法时记录答案

B：3种情况：能减少2，能减少1，不能减少分别考虑清楚

C：利用一个set和一个multiset，把行列分开考虑，利用set自带的排序和查询，每次把相应的块拿出来分成两块插入回去，然后行列分别取最大相乘的作为这次询问的答案

D：一个区间覆盖问题的变形，注意公式的话，很容易发现其实x，w对应的就是一个[x - w, x + w]的区间，然后求最多不重合区间即可

代码：

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

typedef long long ll;

ll a, b, ans;

ll gcd(ll a, ll b) {
    if (!b) return a;
    ans += a / b;
    return gcd(b, a % b);
}

int main() {
    scanf("%lld%lld", &a, &b);
    gcd(a, b);
    printf("%lld\n", ans);
    return 0;
}

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

const int N = 200005;
int n;
char a[N], b[N];

int g[30][30];
int vis[30];

int main() {
    scanf("%d%s%s", &n, a + 1, b + 1);
    int sum = 0;
    for (int i = 1; i <= n; i++)
        if (a[i] != b[i]) sum++;
    for (int i = 1; i <= n; i++) {
        if (a[i] != b[i]) {
            if (g[b[i] - 'a'][a[i] - 'a']) {
                printf("%d\n%d %d\n", sum - 2, i, g[b[i] - 'a'][a[i] - 'a']);
                return 0;
            }
            g[a[i] - 'a'][b[i] - 'a'] = i;
        }
    }
    for (int i = 1; i <= n; i++) {
        if (a[i] != b[i]) {
            vis[b[i] - 'a'] = i;
        }
    }
    for (int i = 1; i <= n; i++) {
        if (a[i] != b[i] && vis[a[i] - 'a']) {
            printf("%d\n%d %d\n", sum - 1, i, vis[a[i] - 'a']);
            return 0;
        }
    }
    printf("%d\n-1 -1\n", sum);
    return 0;
}

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

int w, h, n;
set<int> x[2];
multiset<int> xx[2];
char op[2];
int v;
set<int>::iterator it, l, r;
multiset<int>::iterator tmp;

long long gao(int tp) {
    x[tp].insert(v);
    it = x[tp].find(v);
    l = it; l--; r = it; r++;
    xx[tp].erase(xx[tp].find(*r - *l));
    xx[tp].insert(v - *l);
    xx[tp].insert(*r - v);
    long long ans = 1;
    tmp = xx[tp].end(); tmp--;
    ans *= *tmp;
    tmp = xx[!tp].end(); tmp--;
    ans *= *tmp;
    return ans;
}

int main() {
    scanf("%d%d%d", &w, &h, &n);
    x[0].insert(0); x[0].insert(w);
    x[1].insert(0); x[1].insert(h);
    xx[0].insert(w); xx[1].insert(h);
    while (n--) {
        scanf("%s%d", op, &v);
        printf("%lld\n", gao(op[0] == 'H'));
    }
    return 0;
}

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

const int N = 200005;
const int INF = 0x3f3f3f3f;

struct Seg {
    int l, r;
    Seg() {}
    Seg(int l, int r) {
        this->l = l;
        this->r = r;
    }
} seg[N];

int n, x, w;

bool cmp(Seg a, Seg b) {
    return a.r < b.r;
}

int main() {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%d%d", &x, &w);
        seg[i] = Seg(x - w, x + w);
    }
    sort(seg, seg + n, cmp);
    int ans = 0;
    int r = -INF;
    for (int i = 0; i < n; i++) {
        if (seg[i].l >= r) {
            r = seg[i].r;
            ans++;
        }
    }
    printf("%d\n", ans);
    return 0;
}

Codeforces Round #296 (Div. 2) A B C D

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原文地址：http://blog.csdn.net/accelerator_/article/details/44435629