Codeforces 402A 402B 402C 402D

402A 直接暴力

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <sstream>
#include <string>
#include <cstring>
#include <algorithm>
#include <iostream>
#define maxn 2010
#define INF 0x7fffffff
#define inf 10000000
#define MOD 1000000007
#define ull unsigned long long
#define ll long long
using namespace std;

int main()
{
    int k, a, b, v;
    scanf("%d%d%d%d", &k, &a, &b, &v);
    int ans = 0;
    while(a > 0)
    {
        ans ++;
        if(b >= k-1)
        {
            a -= k*v;
            b = b-k+1;
        }
        else
        {
            a -= (b+1)*v;
            b = 0;
        }
    }
    printf("%d\n", ans);
    return 0;
}

402B

修改尽量少的数字使数列成为公差为k的等差数列 （但是不能有负数）

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <sstream>
#include <string>
#include <cstring>
#include <algorithm>
#include <iostream>
#define maxn 2010
#define INF 0x7fffffff
#define inf 10000000
#define MOD 1000000007
#define ull unsigned long long
#define ll long long
using namespace std;

int num[1010], ans[1010], tem[1010];
int main()
{
    int n, k;
    memset(ans, 0, sizeof(ans));
    scanf("%d%d", &n, &k);
    for(int i = 0; i < n; ++ i)
    {
        scanf("%d", &num[i]);
        tem[i] = num[i] - i*k;
    }
    sort(tem, tem+n);
    int j = 0;
    while(tem[j] <= 0) ++ j;
    int now = tem[j], _max = 0, cnt = 0, cc = tem[j];
    for(int i = j; i < n; ++ i)
    {

        if(now == tem[i]) ++ cnt;
        else
        {
            if(cnt > _max)
            {
                _max = cnt;
                cc = now;
            }
            now = tem[i];
            cnt = 1;
        }
    }
    if(cnt > _max)
    {
        _max = cnt;
        cc = now;
    }
    printf("%d\n", n-_max);
    for(int i = 0; i < n; ++ i)
    {
        if(cc+i*k > num[i]) printf("+ %d %d\n", i+1, cc+i*k-num[i]);
        if(cc+i*k < num[i]) printf("- %d %d\n", i+1, num[i]-cc-i*k);
    }
    return 0;
}

402C 猜了一下 直接枚举一个最有可能的

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <sstream>
#include <string>
#include <cstring>
#include <algorithm>
#include <iostream>
#define maxn 2010
#define INF 0x7fffffff
#define inf 10000000
#define MOD 1000000007
#define ull unsigned long long
#define ll long long
using namespace std;

int main()
{
    int t, n, p;
    scanf("%d", &t);
    while(t --)
    {
        scanf("%d%d", &n, &p);
        int now = 2*n+p, st = 1, k = 1;
        while(now--)
        {
            printf("%d %d\n", st, (st+k)%n == 0 ? n : (st+k)%n);
            ++ st;
            if(st == n+1)
            {
                st = 1;
                ++ k;
            }
        }
    }
    return 0;
}

402D

数学加贪心

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <sstream>
#include <string>
#include <cstring>
#include <algorithm>
#include <iostream>
#define maxn 2010
#define INF 0x7fffffff
#define inf 10000000
#define MOD 1000000007
#define ull unsigned long long
#define ll long long
using namespace std;

int a[5010], cou[5010], pp[100000], m;
bool vis[100000];
struct node
{
    int b[5050];
    bool find(int x)
    {
        for(int i = 0; i < m; ++ i)
            if(b[i] == x) return true;
        return false;
    }
};
node prime;

int fi_pri()
{
    int j = 0;
    memset(vis, 1, sizeof(vis));
    for(ll i = 2; i < 100000; ++ i)
        if(vis[i])
        {
            pp[j++] = i;
            for(ll j = i*i; j < 100000; j += i)
                vis[j] = false;
        }
    return j;
}

int gcd(int r, int l)
{
    return l == 0 ? r : gcd(l, r%l);
}

bool can(int i, int j)
{
    int ans = 0;
    for(int k = 0; k < j; ++ k)
        if((i % pp[k] == 0) && (prime.find(pp[k])))
            while(i % pp[k] == 0)
            {
                -- ans;
                i /= pp[k];
            }
        else
            while(i % pp[k] == 0)
            {
                ++ ans;
                i /= pp[k];
            }
    if(prime.find(i)) --ans;
    if(ans < 0) return true;
    return false;
}

int f(int s, int j)
{
    if(s == 1) return 0;
    int i;
    for(i = 0; i < j; ++ i)
        if(s%pp[i] == 0)
            break;
    if(i != j && prime.find(pp[i]))
        return f(s/pp[i], j) -1;
    if(i != j)
        return f(s/pp[i], j) +1;
    if(prime.find(s))
        return -1;
    return 1;
}

int main()
{
    int asd = fi_pri();

    int n, g;
    scanf("%d%d", &n, &m);
    scanf("%d", &a[0]);
    g = cou[0] = a[0];
    for(int i = 1; i < n; ++ i)
    {
        scanf("%d", &a[i]);
        g = cou[i] = gcd(g, a[i]);
    }
    for(int i = 0; i < m; ++ i)
        scanf("%d", &prime.b[i]);
    for(int i = n-1; i >= 0; -- i)
        if(can(cou[i], asd))
            for(int j = 0; j <= i; ++ j)
            {
                a[j] /= cou[i];
                cou[j] /= cou[i];
            }
    int ans = 0;
    for(int i = 0; i < n; ++ i) ans += f(a[i], asd);
    printf("%d\n", ans);
    return 0;
}