AtCoder Beginner Contest 120 解题报告

标签：lld   res   contest   bridge   com   algo   cto   def   reverse   

为啥最近都没有arc啊...

A - Favorite Sound

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <deque>
#include <map>
#include <set>

#define ll long long
#define inf 0x3f3f3f3f
#define il inline

namespace io {

#define in(a) a = read()
#define out(a) write(a)
#define outn(a) out(a), putchar('\n')

#define I_int ll
inline I_int read() {
    I_int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x * f;
}
char F[200];
inline void write(I_int x) {
    if (x == 0) return (void) (putchar('0'));
    I_int tmp = x > 0 ? x : -x;
    if (x < 0) putchar('-');
    int cnt = 0;
    while (tmp > 0) {
        F[cnt++] = tmp % 10 + '0';
        tmp /= 10;
    }
    while (cnt > 0) putchar(F[--cnt]);
}
#undef I_int

}
using namespace io;

using namespace std;

#define N 100010

int a, b, c;

int main() {
    in(a), in(b), in(c);
    printf("%d\n", min(b / a, c));
}

B - K-th Common Divisor

显然，就是求gcd的第k小的因子。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <deque>
#include <map>
#include <set>

#define ll long long
#define inf 0x3f3f3f3f
#define il inline

namespace io {

#define in(a) a = read()
#define out(a) write(a)
#define outn(a) out(a), putchar('\n')

#define I_int ll
inline I_int read() {
    I_int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x * f;
}
char F[200];
inline void write(I_int x) {
    if (x == 0) return (void) (putchar('0'));
    I_int tmp = x > 0 ? x : -x;
    if (x < 0) putchar('-');
    int cnt = 0;
    while (tmp > 0) {
        F[cnt++] = tmp % 10 + '0';
        tmp /= 10;
    }
    while (cnt > 0) putchar(F[--cnt]);
}
#undef I_int

}
using namespace io;

using namespace std;

#define N 100010

int a, b, k, d[N], tot;

int main() {
    in(a), in(b), in(k);
    int g = __gcd(a, b);
    for(int i = 1; i * i <= g; ++i) {
        if(g % i == 0) {
            d[++tot] = i;
            if(g / i != i) d[++tot] = g / i;
        }
    }
    sort(d+1, d + tot + 1); reverse(d + 1, d + tot + 1);
    printf("%d\n", d[k]);
}

C - Unification

栈的经典题。。。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <deque>
#include <map>
#include <set>

#define ll long long
#define inf 0x3f3f3f3f
#define il inline

namespace io {

#define in(a) a = read()
#define out(a) write(a)
#define outn(a) out(a), putchar('\n')

#define I_int ll
inline I_int read() {
    I_int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x * f;
}
char F[200];
inline void write(I_int x) {
    if (x == 0) return (void) (putchar('0'));
    I_int tmp = x > 0 ? x : -x;
    if (x < 0) putchar('-');
    int cnt = 0;
    while (tmp > 0) {
        F[cnt++] = tmp % 10 + '0';
        tmp /= 10;
    }
    while (cnt > 0) putchar(F[--cnt]);
}
#undef I_int

}
using namespace io;

using namespace std;

#define N 100010
char s[N];
int n, st[N];

int main() {
    scanf("%s", s + 1);
    n = strlen(s + 1);
    int ans = 0, top = 0;
    for(int i = 1; i <= n; ++i) s[i] -= '0';
    for(int i = 1; i <= n; ++i) {
        if(top && s[top] ^ s[i]) {++ans, --top; continue;}
        s[++top] = s[i];
    }
    printf("%d\n", ans * 2);
}

D - Decayed Bridges

正着计数太麻烦了。
考虑最后一条边断掉之后答案肯定是\(n(n-1)/2\)。
用个经典的套路，把删边弄成倒着连边。
那么每次多连一条边，联通的点数就多了\(siz_x*siz_y\)
用\(n(n-1)/2\)减去\(siz_x*siz_y\)即可。
注意longlong。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <deque>
#include <map>
#include <set>

#define ll long long
#define inf 0x3f3f3f3f
#define il inline

namespace io {

#define in(a) a = read()
#define out(a) write(a)
#define outn(a) out(a), putchar('\n')

#define I_int ll
inline I_int read() {
    I_int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x * f;
}
char F[200];
inline void write(I_int x) {
    if (x == 0) return (void) (putchar('0'));
    I_int tmp = x > 0 ? x : -x;
    if (x < 0) putchar('-');
    int cnt = 0;
    while (tmp > 0) {
        F[cnt++] = tmp % 10 + '0';
        tmp /= 10;
    }
    while (cnt > 0) putchar(F[--cnt]);
}
#undef I_int

}
using namespace io;

using namespace std;

#define N 100010

int n, m;
int a[N], b[N], f[N];
ll siz[N];
ll ans = 0, sum = 0, res[N];
int find(int x) {
    if(f[x] == x) return x; 
    else return f[x] = find(f[x]);
}
int main() {
    in(n), in(m);
    sum = (ll)(n * (n - 1ll) / 2ll);
    for(int i = 1; i <= m; ++i) in(a[i]), in(b[i]);
    for(int i = 1; i <= n; ++i) f[i] = i, siz[i] = 1ll;
    for(int i = m; i; --i) {
        res[i] = sum - ans;
        int x = find(a[i]), y = find(b[i]);
        if(x != y) {
            ans += (ll)siz[x] * siz[y];
            siz[x] += siz[y];
            f[y] = x; 
        }
    }
    for(int i = 1; i <= m; ++i) printf("%lld\n", res[i]);
}

AtCoder Beginner Contest 120 解题报告

标签：lld   res   contest   bridge   com   algo   cto   def   reverse   

原文地址：https://www.cnblogs.com/henry-1202/p/10467882.html