Just Say hello to algorithm

-I am so short of practice and so stupid.

Just Say hello to algorithm

Do a little before the simple topic, a is very water, however, WA once, because no greater than 2^32 to open long long, with int not, int range is -2^32~2^32-1

Question F

Design Tutorial:learn from Math

Test instructions: give you a number greater than or equal to 12, you need to use two composite to express it. Composite refers to the number of natural numbers that can be divisible (excluding 0) by other numbers except those that are 1 and the whole.

Analysis: Why give you big and equal to 12, we know even 2 is composite, so give you an even, you subtract an even number to get is even, because n>=12, so minus 4, get is >=8 of even, certainly is composite, of course you can also subtract 6, 8, more than 8 will not be.

And give you an odd number? You subtract an odd number and get an even number, minus one is an odd number of composite, the smallest is 9, odd: n>=13,n-9>=4

#include <stdio.h>int main () {    int A, b;    int n;    scanf ("%d", &n);    if (n%2) {        printf ("%d%d\n", 9,n-9);    } else{        printf ("%d%d\n", 8,n-8);    }    return 0;}

This problem can also be enumerated

#include <stdio.h>int iscomposite (int a) {    if (a==2) return 0;    for (int i=2;i*i<=a;i++)        if (a%i==0) return 1;    return 0;} int main () {    int a,n;    scanf ("%d", &n);    for (a=2;a<n;a++)        if (Iscomposite (a) &&iscomposite (n-a)) {            printf ("%d%d", a,n-a);            break;        }    return 0;}

　　

H Question pseudoprime numbers

Test instructions: Ask 1^m + 2^m + 3^m + ... + n^m (answer mod 1e9+7)

Note: Each time you add up and add up the MoD

#include <stdio.h> #define M 1000000007long long Qpow (Long long A,long long M)//fast power {    long long ans=1,k=a;    while (m)    {        if (m&1)            ans= (ans*k)%M;        k= (k*k)%M;        m>>=1;    }    return ans;} int main () {    long long N,m,ans;    while (scanf ("%lld%lld", &n,&m)!=eof)    {        ans=0;        for (int i=1; i<=n; i++)            Ans+=qpow (i,m)%m;//here mod        a bit printf ("%lld\n", ans%m);//mod here    }    return 0;}

　　

G-Question Swapsort

Test instructions: Using less than n times of swap to arrange the order.

Analysis: Select sort, choose the smallest one after unsorted and then the first interchange that is not sorted, so that a maximum of n-1 can be sorted and the two-digit position of the interchange is recorded in an array.

#include <stdio.h>int n,i,j,a[3002],coun,temp,swap1[3001],swap2[3001];int Main () {    scanf ("%d", &n);    for (i=0; i<n; i++)        scanf ("%d", a+i);    for (i=0; i<n-1; i++)    {        int minj=i;        for (j=i+1; j<n; j + +)            if (A[j]<a[minj]) minj=j;        if (I!=minj)        {            swap1[coun]=i;            Swap2[coun]=minj;            Temp=a[minj];            A[minj]=a[i];            A[i]=temp;            coun++;        }    }    printf ("%d\n", Coun);    if (Coun)        for (i=0; i<coun; i++)            printf ("%d%d\n", Swap1[i],swap2[i]);    return 0;}

　　

-I'm too stupid to refuel!

Just Say hello to algorithm