PE刷题记

PE

中文翻译

最喜欢做这种很有意思的数学题了虽然数学很垃圾

但是这个网站的提交方式好鬼畜啊qwq

1.Multiples of 3 and 5

直接枚举

2.Even Fibonacci numbers

直接枚举

3.Largest prime factor

$sqrt(n)$枚举

#include<cstdio>
#include<vector>
#include<set>
#include<algorithm>
#define sit 
#define LL long long 
using namespace std;
const int MAXN = 1e6 + 10;
inline int read() {
    char c = getchar(); LL x = 0, f = 1;
    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;
}
LL ans = 600851475143;
int main() {
    LL out = 0;
    for(LL i = 2; i * i <= ans; i++) {
        if(ans % i == 0) {
            out = max(out, i);
            while(ans % i == 0) ans /= i;
        }
    }
    printf("%I64d", max(out, ans));
    return 0;
}
/*
*/

4.Largest palindrome product

暴力枚举

#include<cstdio>
#include<vector>
#include<set>
#include<algorithm>
#define sit 
#define LL long long 
using namespace std;
const int MAXN = 1e6 + 10;
inline int read() {
    char c = getchar(); LL x = 0, f = 1;
    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;
}
int ans;
int main() {
    int ans = 0;
    for(int i = 100; i <= 999; i++) {
        for(int j = 100; j <= 999; j++) {
            int x = i * j;
            int a[20], tot = 0, flag = 0;
            while(x) a[++tot] = x % 10, x /= 10;
            for(int k = 1; k <= tot; k++)
                if(a[k] != a[tot - k + 1]) {flag = 1; break;} 
            if(flag == 0) ans = max(ans, i * j);
        }
    }
    printf("%d", ans);
    return 0;
}
/*
*/

5.Smallest multiple

这个就比较有意思了。

首先我们吧所有的数质因数分解，取每个质数的最大指数，乘起来

update：woc？好像求个最小公倍数就行了？？

#include<cstdio>
#include<algorithm>
using namespace std;
int mx[21], prime[21] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71};
void getmax(int x) {
    for(int i = 0; prime[i] <= x && i <= 20; i++) {
        int cur = 0;
        if(x % prime[i] == 0) 
            while(x % prime[i] == 0) 
                cur++, x /= prime[i];
        mx[i] = max(mx[i], cur);
    }
        
}
int main() {
    int N = 20;
    for(int i = 1; i <= N; i++) 
        getmax(i);
    int ans = 1;
    for(int i = 0; i <= N; i++) 
        for(int j = 1; j <= mx[i]; j++)
            ans = ans * prime[i];
    printf("%d", ans);
}

6.Sum square difference

直接枚举

#include<cstdio>
#include<algorithm>
#define LL long long 
using namespace std;
int N = 100;
int main() {
    LL ans = 5050 * 5050;
    for(int i = 1; i <= N; i++) ans = ans - i * i;
    printf("%d", ans);
}

7.10001st prime

直接一波线性筛

#include<cstdio>
#include<algorithm>
#define LL long long 
using namespace std;
const int MAXN = 1e6 + 10;
int N = 1e6;
int prime[MAXN], vis[MAXN], tot = 0;
int main() {
    for(int i = 2; i <= N; i++) {
        if(!vis[i]) prime[++tot] = i;
        for(int j = 1; j <= tot && i * prime[j] <= N; j++) {
            vis[i * prime[j]] = 1;
            if(!(i % prime[j])) break;
        }
    }
    printf("%d", prime[10001]);
}

2018-07-21 07:37:01

8.Largest product in a series

这题目真鬼畜，首先把所有的数copy到一个txt里

然后暴力枚举就行了

2018-07-21 08:01:17

#include<cstdio>
#include<algorithm>
#include<cstring>
#define LL long long 
using namespace std;
int N;
char s[1050];
int main() {
    freopen("a.in", "r", stdin);
    scanf("%s", s + 1);
    int N = strlen(s + 1);
    LL ans = 0;
    for(int i = 1; i <= N; i++) {
        LL now = 1;
        for(int j = i; j <= i + 12 && j <= N; j++) {
            LL x = s[j] - '0';
            now = now * x;
        }
        ans = max(ans, now);
    }
    printf("%I64d", ans);
}

9.Special Pythagorean triplet

大力枚举

#include<cstdio>
#include<algorithm>
#include<cstring>
#define LL long long 
using namespace std;

int main() {
    for(int i = 1; i <= 1000; i++)    
        for(int j = 1; j <= 1000; j++)
            for(int k = 1; k <= 1000; k++) {
                if((i * i + j * j == k * k) && (i + j + k == 1000)) {
                    printf("%d %d %d", i * j * k);
                }
            }
}

10.Summation of primes

大力枚举、、

#include<cstdio>
#include<algorithm>
#include<cstring>
#define LL long long 
using namespace std;
const int MAXN = 1e7 + 10;
int N = 2000001;
LL prime[MAXN], vis[MAXN], tot;
int main() {
    for(int i = 2; i <= N; i++) {
        if(!vis[i]) prime[++tot] = i;
        for(int j = 1; j <= tot && i * prime[j] <= N; j++) {
            vis[i * prime[j]] = 1;
            if(!(i % prime[j])) break;
        }
    }
    for(int i = 1; i <= tot; i++) prime[i] += prime[i - 1];
    printf("%I64d", prime[tot]);
}

2018-07-21 08:08:50

11.Largest product in a grid

直接枚举，一个小优化：考虑到枚举的对称性，每个点都往同一个方向枚举即可

#include<cstdio>
#include<algorithm>
#include<cstring>
#define LL long long 
using namespace std;
const int MAXN = 1e7 + 10;
int N = 20;
LL a[28][28];
int main() {
    LL ans = 0;
    for(int i = 1; i <= N; i++) 
        for(int j = 1; j <= N; j++) 
            scanf("%I64d", &a[i][j]);
    for(int i = 1; i <= N; i++)    
        for(int j = 1; j <= N; j++) {
            ans = max(ans, a[i][j] * a[i + 1][j] * a[i + 2][j] * a[i + 3][j]);
            ans = max(ans, a[i][j] * a[i + 1][j + 1] * a[i + 2][j + 2] * a[i + 3][j + 3]);
            ans = max(ans, a[i][j] * a[i + 1][j - 1] * a[i + 2][j - 2] * a[i + 3][j - 3]);
            printf("%I64d
", ans);
        }
    
}

2018-07-21 08:17:13

12.Highly divisible triangular number

不会做，然后暴力枚举了一发过了qwq。

2018-07-21 09:15:32

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#define LL long long 
#define int long long 
using namespace std;
const int MAXN = 1e6 + 10, INF = 1e9 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    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;
}
int a[MAXN], sum[MAXN];
int prime[MAXN], vis[MAXN], tot = 0;
void GetPrime(int N) {
    vis[1] = 1; prime[0] = 1;
    for(int i = 2; i <= N; i++) {
        if(!vis[i]) prime[++tot] = i;
        for(int j = 1; j <= tot && prime[j] * i <= N; j++) {
            vis[i * prime[j]] = 1;
            if(!i % prime[j]) break;
        }
    }
}
int pd(int val) {
    int ans = 1;
    for(int i = 1; prime[i] <= val; i++) {
        if(val % prime[i] == 0) {
            int now = 0;
            while(val % prime[i] == 0) now++, val /= prime[i];
            ans = ans * (now + 1);
        }
    }
    return ans + (val != 0);
}
main() { 
#ifdef WIN32
    //freopen("a.in", "r", stdin);
#endif
    GetPrime(1e6 + 10);
    a[1] = 1;
    for(int i = 2; i <= 100000; i++) {
        a[i] = a[i - 1] + i;
        sum[i] = pd(a[i]);
        if(sum[i] >= 500) {printf("%I64d", a[i]); return 0;}
        //printf("%d
", sum[i]);
        
    }
    return 0;
}

13.Large sum

直接上高精

2018-07-21 09:56:14

14.Longest Collatz sequence

直接一波记忆化搜索

#pragma comment(linker,"/STACK:1024000000,1024000000")
#include<cstdio>
#include<map>
#define LL long long 
using namespace std;
const int MAXN = 1e6 + 10, INF = 1e9 + 10;
LL N = 1e6;
map<LL, LL> val;
int dfs(LL i) {
    if(val[i]) return val[i];
    if(i & 1) val[i] = dfs(3 * i + 1) + 1;
    else val[i] = dfs(i / 2) + 1;
    return val[i];
}
main() { 
    //freopen("a.in", "r", stdin);
    val[1] = 1;
    int ans = 1, out = 1;
    for(int i = 2; i <= N; i++) {
        int now = dfs(i);
        if(now > ans) ans = now, out = i;
    }
    printf("%d", out);
    return 0;
}

2018-07-21 10:07:51

15.Lattice paths

再一波记忆化搜索。。

#pragma comment(linker,"/STACK:1024000000,1024000000")
#include<cstdio>
#include<map>
#define LL long long 
using namespace std;
const int MAXN = 1e6 + 10, INF = 1e9 + 10;
LL N = 1e6;
LL ans[21][21];
LL dfs(int x, int y) {
    if(ans[x][y]) return ans[x][y];
    if(x - 1 >= 0) ans[x][y] += dfs(x - 1, y); 
    if(y - 1 >= 0) ans[x][y] += dfs(x, y - 1);
    return ans[x][y];
}
main() { 
    //freopen("a.in", "r", stdin);
    ans[0][0] = 1;
    printf("%I64d", dfs(20, 20));
    return 0;
}

2018-07-21 10:14:35

16.Power digit sum

高精度

#pragma comment(linker,"/STACK:1024000000,1024000000")
#include<cstdio>
#include<map>
#define LL long long 
using namespace std;
const int MAXN = 1e6 + 10, INF = 1e9 + 10;
LL N = 1e6;
LL ans[21][21];
LL dfs(int x, int y) {
    if(ans[x][y]) return ans[x][y];
    if(x - 1 >= 0) ans[x][y] += dfs(x - 1, y); 
    if(y - 1 >= 0) ans[x][y] += dfs(x, y - 1);
    return ans[x][y];
}
main() { 
    //freopen("a.in", "r", stdin);
    ans[0][0] = 1;
    printf("%I64d", dfs(20, 20));
    return 0;
}

21.Amicable numbers

暴力。。

#pragma comment(linker,"/STACK:1024000000,1024000000")
#include<cstdio>
#include<map>
#define LL long long 
using namespace std;
const int MAXN = 1e6 + 10, INF = 1e9 + 10;
int vis[MAXN];
int get(int x) {
    int ans = 0;
    for(int i = 1; i < x; i++)
        if(x % i == 0) ans += i;
    return ans;
}
main() { 
    int ans = 0, N = 10000;
    for(int i = 1; i <= N; i++) {
        if(vis[i]) continue;
        int x = get(i), y = get(x);
        //printf("%d %d %d
", i, x, y);
        if(x <= N && y == i && i != x) ans += i + x, vis[i] = 1, vis[x] = 1;
    }
    printf("%d", ans);
    return 0;
}

50.Consecutive prime sum

开始以为有单调性，也就是如果长度为$x$的能构成素数，那$x - 1$一定能构成素数

但是这样枚举是错的qwq。比如$18$不是素数，但是$29$是素数

然后把二分改成枚举就过了。。

2018-07-21 09:15:19

#include<cstdio>
#include<algorithm>
#include<cstring>
#define LL long long 
using namespace std;
const LL MAXN = 1e6 + 10;
LL N = 1000000;
LL prime[MAXN], vis[MAXN], tot = 0;
LL ans = 0;
LL pd(LL num)
{
    if(num==2||num==3) return 1;
    if(num%6!=1&&num%6!=5) return 0;
    for(register LL i=5;i*i<=num;i+=6)
        if(num%i==0||num%(i+2)==0)
            return 0;
    return 1;
}
int len = 0;
bool check(LL num) {
    for(LL i = 1; i <= tot; i++) {
        LL r = prime[i + num - 1], l = prime[i - 1];
        if(r - l > N) return 0;
        if(pd(r - l) == 1 && (i + num - 1 <= tot) && (r - l <= N) && num > len) {
            ans = r - l;
            len = num;
            return 1;
        }
    }
    return 0;
}
int main() {
    for(LL i = 2; i <= N; i++) {
        if(!vis[i]) prime[++tot] = i;
        for(LL j = 1; j <= tot && i * prime[j] <= N; j++) {
            vis[i * prime[j]] = 1;
            if(!(i % prime[j])) break;
        }
    }
    for(LL i = 1; i <= tot; i++) 
        prime[i] += prime[i - 1];
    for(int i = 1; i <= tot; i++)    
        check(i);
    printf("%I64d ", ans);
}