第10問
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.
#include <iostream> #include <math.h> using namespace std; long long int prime[1000000]; long long int table[100000000]; int primeCall(){ int flag=0; prime[0]=2; prime[1]=3; for(long long int i=0;i<3000000;i++) { if(i%2==0)table[i]=0; else table[i]=1; } for(long long int p=1;p<1000000;p++) { if(p%100==0)cout<<prime[p]<<endl; for(long long int i=prime[p];i<2001000;i+=2) { if(i%prime[p]!=0 && flag==0 && table[i]==1) { prime[p+1]=i; flag=1; } if(i%prime[p]==0) { table[i]=0; } } flag=0; if(prime[p+1]>2000000)break; }
8問目。
Find the greatest product of five consecutive digits in the 1000-digit number.
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
#include<iostream> using namespace std; char str[]="7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450"; int main(){ int i=0;//1000桁目、100桁目、10桁目、1桁目をカウントする int num=1; int max=0; for(int p=0;str[p+4]!='\0';p++) { for(int m=10000; m>=1; m/=10,i++) { cout<<(str[i+p]-'0'); num*=(str[i+p]-'0'); } cout<<endl; cout<<num<<endl; if(max < num) max=num; i=0; num=1; } cout<<max<<endl; }
7問目。
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
What is the 10001st prime number?
#include <iostream> using namespace std; long long int primeCall(unsigned int NM){ long long int prime[50000]; long long int *box=new long long int[1000000]; long long int p=0; for(p=0;p<NM;p++)prime[p]=0; prime[0]=2; box[0]=0; for(p=1;p<=NM;p++) { for(long long int i=prime[p-1];i<1000000;i++) { if((i%prime[p-1])==0)box[i]=0; else if (box[i]!=0)box[i]=i; } for(long long int i=2;i<1000000;i++) { if(box[i]!=0) { prime[p]=i; break; } } } delete [] box; return prime[NM-1]; } int main() { cout<<primeCall(10001)<<endl; }
6問目。
The sum of the squares of the first ten natural numbers is,
1^2 + 2^2 + ... + 10^2 = 385
The square of the sum of the first ten natural numbers is,(1 + 2 + ... + 10)^2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 385 = 2640.Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
#include <iostream> using namespace std; int main() { unsigned long long int sum=0; unsigned long long int sq=0; for(int i=1;i<=100;i++) { sum+=(i*i); sq+=i; } cout<<(sq*sq)-sum<<endl; }
5問目。
2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
What is the smallest number that is evenly divisible by all of the numbers from 1 to 20?
#include <iostream> using namespace std; int main() { for(int i=2520;i<=1000000000;i++) { if( (i%11)==0 && (i%12)==0 && (i%13)==0 && (i%14)==0 && (i%15)==0 && (i%16)==0 && (i%17)==0 && (i%18)==0 && (i%19)==0 &&(i%20)==0 )cout<<i<<endl; } }
4問目。
A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91*99.
Find the largest palindrome made from the product of two 3-digit numbers.
#include <iostream> using namespace std; int main() { int max=0; for(int i=100;i<=999;i++) { for(int n=100;n<=999;n++) { max=(i*1000)+(i/100)+((i/10%10)*10)+((i%10)*100); if((max % n)==0 && (max/ n) > 100 && (max/ n) < 1000) { cout<<n<<endl; cout<<max/n<<endl; cout<<max<<endl; break; } } }
第2問
#include <iostream> #include <time.h> using namespace std; int main() { unsigned long long int l=0; unsigned long long int r=1; unsigned long long int m=0; unsigned long int ans =0; for(long long int i=0;i<=4000000;i++){ if(m>4000000)break; m=l+r; //cout<<m<<endl; if(m%2==0)ans=ans+m; l=r; r=m; } cout<<ans<<endl; }