Project Euler 1-25

A. Multiples of 3 and 5

大水题，计算出below n内是3，5倍数的总和

数据较大，只能用求和公式

#include <iostream>
#include <cstdio>
typedef long long ll;
using namespace std;int main()
{int T;scanf("%d",&T);while(T--){int n;scanf("%d",&n);ll sum=0;ll p;p = (n-1)/3;sum += (1LL)*3*(p+1)*p/2;p = (n-1)/5;sum += (1LL)*5*(p+1)*p/2;p = (n-1)/15;sum -= (1LL)*15*(p+1)*p/2;cout<<sum<<endl;}//cout << "Hello world!" << endl;return 0;
}

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B.Even Fibonacci numbers

求出n以内所有偶数的总和

所有偶数都是3个一循环出现，序列为0，2，8，34，144...

发现规律E【i】 = 4*E【i-1】+E【i-2】

打表计算

#include <iostream>
#include <cstdio>
typedef long long  ll;
using namespace std;
const int MAXN = 1e3+100;
const ll MAXNN = 4e16;ll a[MAXN];
ll sum[MAXN];
ll top;int main()
{a[0] = 0;a[1] = 2;sum[0] = 0;sum[1] = 2;for(ll i=2;;i++){a[i] = 4*a[i-1]+a[i-2];top ++;sum[i] = sum[i-1]+a[i];//cout<<top<<endl;if(a[i]>MAXNN)break;}top+=2;//cout<<a[2]<<a[3]<<" "<<a[4]<<endl;int T;scanf("%d",&T);while(T--){ll n;scanf("%lld",&n);int l=0,r=top;while(l<r){int m = l+(r-l)/2;if(a[m]<=n)l = m+1;else r = m;}printf("%lld\n",sum[l-1]);}//cout << "Hello world!" << endl;return 0;
}

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http://digi.tech.qq.com/a/20150526/008620.htm

我发现这个fib时钟挺好玩的

时钟:红色+蓝色

分钟:(绿色+蓝色)*5

C. Largest prime factor

求最大质因子

prime打表，昨天打了bc，去求了最大因子，呵呵

#include <iostream>
#include <cstdio>
typedef long long  ll;
using namespace std;
const int MAXN = 1e7+1000;
//const ll MAXNN = 4e16;int visit[MAXN];
ll p[MAXN];
ll cnt;void isprime(){cnt = 0;for(ll i=2;i<MAXN;i++){if(!visit[i]){p[cnt++] = i;for(ll j=i*2;j<MAXN;j+=i){visit[j] = 1;}}}
}int main()
{isprime();//cout<<p[1]<<endl;//cout<<p[2]<<endl;int T;scanf("%d",&T);while(T--){ll n;scanf("%lld",&n);// cout<<n<<endl;ll mmax = -1;for(ll i=0;i<cnt;i++){if(n%p[i]==0){mmax = max(mmax,p[i]);while(n%p[i]==0)n/=p[i];}}if(n>1){mmax = max(mmax,n);}cout<<mmax<<endl;// cout<<n<<endl;
}return 0;
}

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H.Largest product in a series

求k个连续序列最大值

遍历的时候考虑前一个序列的第一个值是否为0

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
using namespace std;const int MAXN = 1e3+100;
char ts[MAXN];int main(){int T;scanf("%d",&T);while(T--){int n,k;scanf("%d%d",&n,&k);scanf("%s",ts);int len = strlen(ts);int mmax=1;for(int i=0;i<k;i++)mmax *= (ts[i]-'0');int pre = mmax;//cout<<pre<<endl;for(int i=k;i<len;i++){//pre = pre/(ts[i-k]-'0')*(ts[i]-'0');if(ts[i-k]-'0'==0){pre = 1;int s = i-k+1;for(int j=1;j<=k;j++){pre *= (ts[s++]-'0');}}else{pre = pre*(ts[i]-'0')/(ts[i-k]-'0');}mmax = max(mmax,pre);//cout<<pre<<endl;
        }cout<<mmax<<endl;}return 0;
}

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I.Special Pythagorean triplet

求构成勾股数的abc，乘积最大

一眼看成求最小，测试数据能过，又傻逼了，改了一个小时看不出，差点要哭

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
typedef long long ll;
using namespace std;const ll MAXN = 3e3+100;
ll p[MAXN];int main()
{//memset(p,-1,sizeof p);for(int i=1;i<MAXN/3;i++){for(int j=i+1;j<MAXN/2;j++){ll t = (1LL)*i*i + (1LL)*j*j;ll x = sqrt(t);if(t==x*x&&j<x&&i+j+x<MAXN){if(p[i+j+x]==0)p[i+j+x] = (1LL)*i*j*x;elsep[i+j+x] = max(p[i+j+x],(1LL)*i*j*x);}}}int T;scanf("%d",&T);while(T--){int n;scanf("%d",&n);if(p[n]==0){cout<<-1<<endl;continue;}cout<<p[n]<<endl;}//cout << "Hello world!" << endl;return 0;
}

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L.Highly divisible triangular number

这个挺坑的，反正思路挺清晰的，n的所有因子总数是 (1+a1)*(1+a2)*(1+a3)...(ai为n的质因子)

至于为什么坑了那么久，就是因为打表的时候，我数组我不敢开太大，orz

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>typedef long long ll;using namespace std;const int MAXN = 1e4;
const int MAXNN = 1e6;
int vis[MAXN];
int p[MAXN];
int cnt;
ll s[MAXNN];
ll num[MAXNN];
int top;void isprime(){cnt= 0;for(int i=2;i<MAXN;i++){if(!vis[i]){p[cnt++] = i;for(int j=2*i;j<MAXN;j+=i){vis[j] = 1;}}}
}int main()
{isprime();top = 1;for(int i=1;i<MAXNN;i++)s[i] = s[i-1]+i;//cout<<s[MAXNN-1]<<endl;for(int i=1;i<MAXNN;i++){ll t = s[i];ll k = 0;ll sum=1;for(int j=0;j<cnt&&p[j]<=sqrt(t);j++){k=0;if(t%p[j]==0){while(t%p[j]==0){t/=p[j];k++;}sum *= (k+1);}}if(t>1)sum *= 2;num[i] = sum;top++;if(sum>1000){break;}//cout<<sum<<endl;
    }// cout<<top<<endl;//cout<<num[top-1]<<endl;//cout<<s[top-1]<<endl;int T;scanf("%d",&T);while(T--){int n;scanf("%d",&n);ll ans=-1;for(int i=1;i<=top;i++){if(num[i]>n){ans = s[i];//cout<<i<<endl;break;}}cout<<ans<<endl;}//cout << "Hello world!" << endl;return 0;
}

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M.Large sum

大数模板

http://blog.csdn.net/vsooda/article/details/8543351

http://blog.csdn.net/y990041769/article/details/20116995

这些模板都没负数操作的样子。。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>using namespace std;string sum(string s1,string s2)
{if(s1.length()<s2.length()){string temp=s1;s1=s2;s2=temp;}int i,j;for(i=s1.length()-1,j=s2.length()-1;i>=0;i--,j--){s1[i]=char(s1[i]+(j>=0?s2[j]-'0':0));   //注意细节if(s1[i]-'0'>=10){s1[i]=char((s1[i]-'0')%10+'0');if(i) s1[i-1]++;else s1='1'+s1;}}return s1;
}int main()
{int n;scanf("%d",&n);string ts1;cin>>ts1;for(int i=1;i<n;i++){string ts2;cin>>ts2;ts1 = sum(ts1,ts2);}for(int i=0;i<10;i++)cout<<ts1[i];cout<<endl;//cout << "Hello world!" << endl;return 0;
}

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N.Longest Collatz sequence

这题很坑

n可能大于int了(奇数为3*n+1)，所以这就要开longlong

一天都不知道错在哪里

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
using namespace std;const int MAXN = 5e6+10;
typedef long long ll;
int a[MAXN],b[MAXN];int dfs(ll x){if(x<MAXN&&a[x]){return a[x];}int sum = 0;if(x%2==0){sum = dfs(x/2)+1;}else{sum = dfs(x*3+1)+1;}if(x<MAXN)a[x] = sum;return sum;
}int main()
{a[1]=b[1]=1;int mmax = -1;int pre = 1;for(int i=2;i<MAXN;i++){if(!a[i])a[i] = dfs(i);if(mmax<=a[i]){pre = i;mmax = a[i];}b[i] = pre;}int T;scanf("%d",&T);while(T--){int n;scanf("%d",&n);printf("%d\n",b[n]);}//cout << "Hello world!" << endl;return 0;
}

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O.Lattice paths

简单dp

#include <cstdlib>
#include <cstring>
#include <iostream>
#include <cmath>
#include <cstdio>using namespace std;typedef long long ll;
const ll mod = 1e9+7;
const int MAXN = 5e2+10;ll dp[MAXN][MAXN];int main(){for(int i=1;i<MAXN;i++){for(int j=1;j<MAXN;j++){if(j==1&&i==1)dp[i][j] = 1;elsedp[i][j] = dp[i-1][j]+dp[i][j-1];dp[i][j] %= mod;}}int T;scanf("%d",&T);while(T--){int n,m;scanf("%d%d",&n,&m);cout<<dp[n+1][m+1]<<endl;}
}

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P.Power digit sum

字符串操作

#include <cstdlib>
#include <cstring>
#include <iostream>
#include <cmath>
#include <cstdio>using namespace std;typedef long long ll;
const ll mod = 1e9+7;
const int MAXN = 1e4+100;string ts = "1";ll a[MAXN];int main(){for(int i=1;i<MAXN;i++){int len = ts.length();int flag = 0;ll sum = 0;for(int j=len-1;j>=0;j--){int t = ts[j]-'0';t = t*2+flag;ts[j] = char(t%10+'0');flag = t/10;sum += t%10;}if(flag){ts = "1"+ts;sum += 1;}//cout<<ts<<endl;a[i] = sum;}cout<<ts.length()<<endl;cout<<ts<<endl;int T;scanf("%d",&T);while(T--){int n;scanf("%d",&n);cout<<a[n]<<endl;}
}

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R.Maximum path sum I

简单dp

S.Counting Sundays

计算任意任意区间每月第一天为周七的个数

和省赛一题一模一样，不过好像是求周一的个数，然后我也想当然了，然后第二组测试数据直接出不来。。。

不过学了一个很有用的计算任意时间为周几的蔡勒公式

http://blog.csdn.net/kumxp/article/details/5185309

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;typedef long long ll;/*计算任何日期为周几模板*/
int fun(int x,int y,int z){if(y<3){x-=1;y+=12;}int a = x/100,b = x-100*a;int w = a/4 - 2*a+b+b/4+(13*(y+1)/5)+z-1;w = (w%7+7)%7;return w;
}int main()
{/*for(int i=1;i<13;i++){cout<<fun(2016,i,1)<<" ";}cout<<endl;*/int T;scanf("%d",&T);while(T--){ll tx1;int x1,y1,z1;ll tx2;int x2,y2,z2;scanf("%lld%d%d",&tx1,&y1,&z1);scanf("%lld%d%d",&tx2,&y2,&z2);x1 = (tx1-1900)%400+1900;x2 = (tx2-tx1)+x1;int ans = 0;for(int i=x1;i<=x2;i++){int ret = 0;int s = 1,e = 13;if(i==x1){s = y1;if(z1>1)s++;}if(i==x2){e = y2+1;}for(int j=s;j<e;j++){if(fun(i,j,1)==0)ret++;}//cout<<ret<<endl;ans += ret;}cout<<ans<<endl;}//cout << "Hello world!" << endl;return 0;
}

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T.Factorial digit sum

继续套大数相乘模板

U.Amicable numbers

打表的时候注意是否会超出得到的数字是否会超出数组范围

第二次犯这个错误了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <algorithm>
using namespace std;
typedef long long ll;
const int MAXN = 1e5+1000;
int a[MAXN];
ll b[MAXN];
int main()
{//ll sum = 0;for(int i=1;i<MAXN;i++){int t = i;ll cnt = 0;for(int j=2;j<=sqrt(t);j++){if(t%j==0){cnt += j;if(j*j!=t)cnt+= t/j;}}cnt++;b[i] = cnt;if(cnt<i&&b[cnt]==i){a[i] = a[cnt] = 1;}else if(cnt>MAXN){ll k = 0;for(int j=2;j<=sqrt(k);j++){if(cnt%j==0){k += j;if(j*j!=cnt)k+= cnt/j;}}k++;if(cnt==k)a[i] = 1;}//cout<<sum<<endl;
    }//cout<<b[MAXN-20]<<endl;int T;scanf("%d",&T);while(T--){int n;scanf("%d",&n);ll ret = 0;for(int i=1;i<=n;i++){if(a[i])ret+= i;}cout<<ret<<endl;}return 0;
}

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X.Lexicographic permutations

粗略学习了康托展开

排列组合好像非常有用的样子

http://www.cnblogs.com/xianglan/archive/2011/03/07/1976431.html

http://blog.csdn.net/zhongkeli/article/details/6966805

#include <cstdio>
#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
using namespace std;typedef long long ll;
ll fac[20];
int sg[20];int main(){ll mult = 1;for(int i=1;i<13;i++){mult *= i;fac[12-i] = mult;}int T;string ts;scanf("%d",&T);while(T--){ts = "abcdefghijklm";memset(sg,0,sizeof sg);ll n;scanf("%lld",&n);n--;for(int i=0;i<12;i++){sg[i] = n/fac[i];n -= sg[i]*fac[i];}/*for(int i=0;i<12;i++){cout<<sg[i]<<" ";}cout<<endl;*/for(int i=0;i<12;i++){swap(ts[i],ts[i+sg[i]]);sort(ts.begin()+i+1,ts.end());}cout<<ts<<endl;}
}

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Y.N-digit Fibonacci number

求第一次出现5000-digit fibonacci的下标

这题应该能用字符串的大数操作的样子

学习了一种公式的方法

fib数列如果n较大时,满足an/an+1 = 0.618:1

则an*phi = an,phi=(1+sqrt(5))/2

ox = (1-sqrt(5))/2

可以推出an的通式 an = 1/(sqrt(5))*(phi-ox)^n

则len_dig = (int)(nlog10(phi)-log10(5)/2+1)

呵呵，继续学习了fib

#include <cstdio>
#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;typedef long long ll;
const int MAXN = 5e3+10;
const int MAXNN = 3e4+100;
double golden = (1+sqrt(5))/2;
int a[MAXN];
int len_fib(ll n){return (int)(n*log10(golden) - log10(5)/2+1);
}int main(){//cout<<len_fib(17)<<endl;//ll k = 0;a[1] = 1;for(int i=2;i<MAXN;i++){int l = 1,r = MAXNN;while(l<r){int m = l+(r-l)/2;if(len_fib(m)<i)l = m+1;else r = m;}a[i] = l;}/*cout<<a[2]<<endl;cout<<a[3]<<endl;cout<<a[5000]<<endl;cout<<k<<endl;*/int T;scanf("%d",&T);while(T--){int n;scanf("%d",&n);cout<<a[n]<<endl;}return 0;
}

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转载于:https://www.cnblogs.com/EdsonLin/p/5702123.html