Rokua number theory {water problem} set

1. Sequencing of P1327 sequences

Title Description

Given a sequence {an}, this sequence satisfies Ai≠aj (I≠J), and now requires you to order this sequence from small to large, each time you allow you to exchange any one of them, how many times do I need to exchange?

input/output format

Input format:

The first line, positive integer n (n<=100,000).

Following several lines, total n number, separated by a space, indicating the sequence {an}, arbitrary -231<ai<231.

Output format:

Only one row, containing a number, represents the minimum number of interchanges.

input/Output sampleInput Sample # #:

88 23 4 16 77-5 53 100

Sample # # of output:

 5  
 
  
    I thought it was in reverse order.    
    Find a swap chain, move from the original position to the new location, and end up with a ring    
   Span style= "FONT-SIZE:16PX; Color: #0000ff; " > Note every time you add a cnt-1, because the last two are in the correct place    
    code to implement the original location package structure , sort, flag represents a return, go on.    

#include <iostream> #include <cstdio> #include <algorithm>using namespace std;const int n=100005; struct node{    int o,w;} A[n];bool CMP (const node &A,CONST node &b) {    return A.W<B.W;} int n,flag[n],ans=0;    Guiweiint Main () {    scanf ("%d", &n);    for (int i=1;i<=n;i++) {        scanf ("%d", &a[i].w); a[i].o=i;    }    Sort (a+1,a+1+n,cmp);    for (int i=1;i<=n;i++) {        if (flag[i]==0) {            int now=i,cnt=0;            while (flag[now]==0) {                cnt++;                Flag[now]=1;                NOW=A[NOW].O;            }            ans+=cnt-1;        }    }    Cout<<ans;}

 2.p1630 Summation

Title Description

The remainder of the 1^b+2^b+......+a^b and divided by 10000.

input/output format

Input format:

The first line contains a positive integer n, representing a total of n groups of test data;

Next n rows, each row contains two positive integers a and B.

"Data Size"

For 30% of the data, meet the n<=10,a,b<=1000;

For 100% of the data, meet the n<=100,a,b<=1000000000;

Output format:

A total of n rows, one corresponding answer per line.

input/Output sampleInput Sample # #:

12 3

Sample # # of output:

 9  
 -- ---------------------------------------------------------------------------------------- 
    violence can only take 30 points    
    use F[i] to represent the results of i^b%10000, consider each fi contribution to the answer    
    complexity O (10000*logb*n)    
   do not use long long can also   

#include <iostream> #include <cstdio> #include <algorithm> #include <cstring>using namespace Std;const int mod=10000,n=10005;typedef Long long ll;int n;ll a,b;ll f[n];ll powmod (ll a,ll b) {    ll ans=1;    for (; b;b>>=1,a= (a*a)%mod)        if (b&1) ans= (ans*a)%mod;    return ans;} ll solve (ll A,ll b) {for    (int i=1;i<=mod;i++)        f[i]=powmod (i,b);    ll Div=a/mod,mod=a%mod;    ll Ans=0;    for (int i=1;i<=mod;i++) {        ans= (ans+div*f[i])%mod;        if (i<=mod) ans= (ans+f[i])%mod;    }    return ans;} int main () {    scanf ("%d", &n);    for (int i=1;i<=n;i++) {        memset (f,0,sizeof (f));        scanf ("%lld%lld", &a,&b);        Cout<<solve (A, b) << "\ n";    }}

An approximate study of 3.p1403

Title Description

The scientists ' exploration of the Samuel Planet has been enriched with energy reserves, making it possible for the long-time operations of large computers in the space station to be "Samuel2". As a result of the hard work of last year, Xiao Lian allowed to use "Samuel2" for mathematical research.

In a recent study of the number of issues related to the sum, he counted the number of approximate numbers of each positive n and expressed it in F (n). Now Xiao Lian hopes to use "Samuel2" to count the sum of f (1) to F (N) and M.

F (n) represents an approximate number of N, which is now given N, which requires the sum of f (1) to F (n) to be calculated.

input/output format

Input format:

Enter a line, an integer n

Output format:

Outputs an integer representing the sum

input/Output sampleInput Sample # #:

3

Sample # # of output:

5

   ----- -------------------------------
  I started with a unique decomposition to add about a few numbers and sums 
   See is to ask for 1~n, can be like the Sieve method, suddenly think, consider the contribution of each number is n/i, too water, ━━ (￣ー￣*| | | ━━ 
     

#include <iostream>usingnamespace  std; int n,ans=0; int Main () {    cin>>N;      for (int i=1; i<=n;i++) ans+=n/i;    cout<<

Rokua number theory {water problem} set