Codeforces 402A 402B 402C 402D

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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 longusing 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 longusing 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 longusing 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 longusing 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;}

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