codeforces-1166 (div2)

最新推荐文章于 2024-05-15 16:42:44 发布

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最新推荐文章于 2024-05-15 16:42:44 发布

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原文链接：http://www.cnblogs.com/Hugh-Locke/p/10946675.html

博客分享了数学题的解题思路，包括均匀分配字母、5 * 5及以上矩阵的构造方法，还介绍了推式子时同正同负和不同正负情况的处理，以及序列中数上下界的确定和贪心算法的运用等内容。

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所以说，遇到数学题就会翻车

A.均匀分配相同的字母就可以了。

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
int read(){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;}
const double eps = 1e-9;
const int maxn = 110;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,M,K;
char c[100000];
int vis[maxn];
int main(){
    Sca(N);
    int ans = 0;
    for(int i = 1; i <= N ; i ++){
        scanf("%s",c);
        int id = c[0] - 'a';
        vis[id]++;
    } 
    for(int i = 0 ; i < 26; i ++){
        int p = vis[i] / 2;
        int q = vis[i] - p;
        ans += p * (p - 1) / 2;
        ans += q * (q - 1) / 2;
    }
    Pri(ans);
    return 0;
}

B.5 * 5及以上的矩阵必定合理,具体构造方法就是

aeiou

eioua

iouae

ouaei

uaeio

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
int read(){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;}
const double eps = 1e-9;
const int maxn = 110;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,M,K;
char c[100000];
int vis[maxn];
const char A[5] = {'a','e','i','o','u'};
int main(){
    Sca(N);
    if(N < 25){
        puts("-1");
        return 0;
    } 
    int x,y;
    for(int i = 5; i <= N; i ++){
        if(!(N % i)){
            x = i;
            break;
        }
    }
    y = N / x;
    if(y < 5 || x < 5){
        puts("-1");
        return 0;
    }
    for(int i = 0; i < x; i ++){
        int id = i % 5;
        for(int j = 0 ; j < y; j ++){
            printf("%c",A[(id + j) % 5]);
        }
    }
    return 0;
}

C.推式子，同正同负的情况很好推，假设a < b && a * b > 0,对于任意a,范围[a,2a]内的所有b是符合条件的，二分查找出来就可以了。

然后是不同正负的情况, a * b < 0的情况,对于任意abs(a) < abs(b),abs(b)范围[a,2a]内的也是符合条件的，同样二分查找出来

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
int read(){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;}
const double eps = 1e-9;
const int maxn = 2e5 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,M,K;
int a[maxn];
int b[maxn];
LL cul(int *x,int *y,int n,int m,int flag){
    LL ans = 0;
    for(int i = 1; i <= n; i ++){
        int p = lower_bound(x + 1,x + 1 + n,x[i] * 2 + 1) - x - 1;
        int l = lower_bound(y + 1,y + 1 + m,x[i] * 2 + 1) - lower_bound(y + 1,y + 1 + m,x[i] + flag);
        ans += p - i + l;
    }
    return ans;
}
int main(){
    Sca(N);
    int cnt1 = 0,cnt2 = 0;
    for(int i = 1; i <= N ; i ++){
        int x = read();
        if(x >= 0) a[++cnt1] = x;
        else b[++cnt2] = -x;
    }
    sort(a + 1,a + 1 + cnt1);
    sort(b + 1,b + 1 + cnt2);
    LL ans = 0;
    ans += cul(a,b,cnt1,cnt2,0);
    ans += cul(b,a,cnt2,cnt1,1);
    Prl(ans);
    return 0;
}

View Code

D.显然序列中每个数的上下界是可以确定的，Mini = 2 ^ (i - 1) * (L + 1),Maxi = 2 ^ (i - 1) * (R + 1)

存在范围内的数必有解，知道了之后最终序列长度也可以确定了。

然后贪心的先取最小的，即所有r[i]等于1，然后贪心的将i增加

ri 需要满足 ri <= R / two[len - i - 1] - 4 * Min[i - 1] + 2 * r[i - 1] - 1

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
int read(){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;}
const double eps = 1e-9;
const int maxn = 110;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,K;
LL L,R,M;
LL a[maxn];
LL Max[maxn],two[maxn],Min[maxn],r[maxn];
int main(){
    Sca(N);
    two[0] = 1;
    for(int i = 1; i <= 60; i ++) two[i] = two[i - 1] * 2;
    while(N--){
        scanf("%lld%lld%lld",&L,&R,&M);
        if(L == R){
            printf("1 %lld\n",L);
            continue;
        }
        int cnt = 2;
        Max[1] = Min[1] = L;
        int len = 0;
        for(int i = 2; i <= 50; i ++){
            Min[i] = two[i - 2] * (L + 1);
            Max[i] = two[i - 2] * (L + M);
            if(Min[i] > R) break;
            if(Min[i] <= R && Max[i] >= R){
                len = i;
                break;
            }
        }
        if(!len){
            puts("-1");
            continue;
        }
        for(int i = 1; i <= len; i ++) r[i] = 1;
        Min[1] = L; r[1] = L;
        for(int i = 2; i < len ; i ++){
            r[i] = min(R / two[len - i - 1] - 4 * Min[i - 1] + 2 * r[i - 1] - 1,M);
            if(i != 2)Min[i] = 2 * Min[i - 1] + r[i] - r[i - 1];
            else Min[i] = L + r[i];
        }
        Prl(len);
        printf("%lld",L);
        Min[len] = R;
        for(int i = 2; i <= len ; i ++){
            printf(" %lld",Min[i]);
        }
        puts("");
    }
    return 0;
}