usaco Milking Cows solution

usaco Milking Cows solution

/*
ID: garakch1
LANG: C++
PROB: milk2
*/
#include <cstdio>
#include <vector>
#include <algorithm>

bool cmp(const std::pair<int, int> &left, const std::pair<int, int> &right) {
  return left.first < right.first;
}

int main() {
  freopen("milk2.in", "r", stdin);
  freopen("milk2.out", "w", stdout);
  std::vector< std::pair<int, int> > v;
  int n, a, b;
  scanf("%d", &n);
  for (int i=0; i<n; i++) {
    scanf("%d%d", &a, &b); 
    v.push_back(std::make_pair(a, b));
  }
  sort(v.begin(), v.end(), cmp);
  //for (int i=0; i<n; i++) printf("%d %d\n", v[i].first, v[i].second);
  int left=v[0].first, right=v[0].second, idle=0, cont;
  cont=right-left;
  for(int i=0; i<n; i++) {
    if(v[i].first>right) idle=std::max(v[i].first-right, idle), left=v[i].first, right=v[i].second;
    else if(v[i].second>right) right=v[i].second, cont=std::max(right-left, cont);
  }
  printf("%d %d\n", cont, idle);
  return 0;
}

usaco Broken Necklace solution

usaco Broken Necklace:

Broken Necklace

You have a necklace of N red, white, or blue beads (3<=N<=350) some of which are red, others blue, and others white, arranged at random. Here are two examples for n=29:

                1 2                               1 2
            r b b r                           b r r b
          r         b                       b         b
         r           r                     b           r
        r             r                   w             r
       b               r                 w               w
      b                 b               r                 r
      b                 b               b                 b
      b                 b               r                 b
       r               r                 b               r
        b             r                   r             r
         b           r                     r           r
           r       r                         r       b
             r b r                             r r w
            Figure A                         Figure B
                        r red bead
                        b blue bead
                        w white bead

The beads considered first and second in the text that follows have been marked in the picture.
The configuration in Figure A may be represented as a string of b's and r's, where b represents a blue bead and r represents a red one, as follows: brbrrrbbbrrrrrbrrbbrbbbbrrrrb .
Suppose you are to break the necklace at some point, lay it out straight, and then collect beads of the same color from one end until you reach a bead of a different color, and do the same for the other end (which might not be of the same color as the beads collected before this).
Determine the point where the necklace should be broken so that the most number of beads can be collected.

Example

For example, for the necklace in Figure A, 8 beads can be collected, with the breaking point either between bead 9 and bead 10 or else between bead 24 and bead 25.
In some necklaces, white beads had been included as shown in Figure B above. When collecting beads, a white bead that is encountered may be treated as either red or blue and then painted with the desired color. The string that represents this configuration can include any of the three symbols r, b and w.
Write a program to determine the largest number of beads that can be collected from a supplied necklace.

PROGRAM NAME: beads

INPUT FORMAT

Line 1:	N, the number of beads
Line 2:	a string of N characters, each of which is r, b, or w

SAMPLE INPUT (file beads.in)

29
wwwbbrwrbrbrrbrbrwrwwrbwrwrrb

OUTPUT FORMAT

A single line containing the maximum of number of beads that can be collected from the supplied necklace.

SAMPLE OUTPUT (file beads.out)

11

OUTPUT EXPLANATION

Consider two copies of the beads (kind of like being able to runaround the ends). The string of 11 is marked.

                Two necklace copies joined here
                             v
wwwbbrwrbrbrrbrbrwrwwrbwrwrrb|wwwbbrwrbrbrrbrbrwrwwrbwrwrrb
                       ******|*****
                       rrrrrb bbbbb  <-- assignments
                       rrrrr#bbbbbb  
                       5 x r  6 x b  <-- 11 total

usaco Broken Necklace solution: http://ideone.com/zdNlhI

/*
ID: [your ID here]
LANG: C++
PROB: beads
*/
#include <iostream>
#include <cstdio>
#include <algorithm>

int n;
char s[355];

int md(int j) {
    while(j<0) j+=n;
    return j%n;
}

int bp(int pos, int dir) {
    int i, j;
    char col='w';
    if(dir==1) j=pos;
    else j=pos-1;
    for(i=0; i<n; i++, j=md(j+dir)) {
        if(col=='w' && s[j]!='w') col=s[j];
        if(col!='w' && s[j]!='w' && s[j]!=col) break;
    }
    return i;
}

int main() {
    freopen("beads.in", "r", stdin);
    freopen("beads.out", "w", stdout);
    int ma=0;
    scanf("%d", &n);
    scanf("%s", s);
    for(int i=0; i<n; i++) ma=std::max(bp(i, -1)+bp(i, 1), ma);
    if(ma>n) ma=n;
    printf("%d\n", ma);
    return 0;
}

alternative solutions for usaco Broken Necklace: http://usacotraining.blogspot.com.tr/2013/09/problem-114-broken-necklace.html

usaco Friday the Thirteenth solution

usaco Friday the Thirteenth:

usaco Friday the Thirteenth solution: http://ideone.com/ClCFT6

#include <iostream>
#include <cstdio>
using namespace std;

int main() {
    int n, l=1, w[10]={0};
    scanf("%d", &n);
    for(int i=1900; i<1900+n; i++) {
        for(int j=0; j<12; j++) {
            if(j==3 || j==5 || j==8 || j==10) {
                for(int k=0; k<30; k++) {
                    l++, l%=7;
                    if(k==12) w[l]++;
                }
            }
            else if(j!=1) {
                for(int k=0; k<31; k++) {
                    l++, l%=7;
                    if(k==12) w[l]++;
                }
            }
            else {
                if(i%400==0 || (i%100!=0 && i%4==0)) {
                    for(int k=0; k<29; k++) {
                        l++, l%=7;
                        if(k==12) w[l]++;
                    }
                }
                else {
                    for(int k=0; k<28; k++) {
                        l++, l%=7;
                        if(k==12) w[l]++;
                    }
                }
            }
        }
    }
    for(int i=0; i<6; i++) printf("%d ", w[i]);
    printf("%d\n", w[6]);
    return 0;
}

note: dont forget to add your usacogate ID, PROG, LANG comment at the beginning of the code. and freopen() stuff in main() part.

usaco Greedy gift givers solution

usaco Greedy gift givers:

usaco Greedy gift givers solution: http://ideone.com/49AVHb

/*
ID: [your usacogate ID here]
PROG: gift1
LANG: C++
*/
#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    freopen("gift1.in", "r", stdin);
    freopen("gift1.out", "w", stdout);
    int n, a, b, k=0, acc[15]={0}, idx, tmp;
    vector <string> v;
    string s;
    scanf("%d", &n);
    for(int i=0; i<n; i++) cin>>s, v.push_back(s);
    for(int i=0; i<n; i++) {
        cin>>s;
        scanf("%d%d", &a, &b);
        idx=find(v.begin(), v.end(), s)-v.begin();
        tmp=idx;
        for(int j=0; j<b; j++) {
            cin>>s;
            idx=find(v.begin(), v.end(), s)-v.begin();
            acc[idx]+=a/b, acc[tmp]-=a/b;
        }
    }
    for(vector <string>::iterator i=v.begin(); i!=v.end(); i++) {
        cout<<(*i)<<" "<<acc[k]<<endl;
        k++;
    }
    return 0;
}

note: function find() belongs to <algorithm> C++ library, and it helps you to find position/index of sepcific thing in C++ vectors. for ref: http://stackoverflow.com/a/15099748/2948746

usaco Your ride is here solution

usaco Your ride is here:

usaco Your ride is here solution: http://ideone.com/GmGx8J

/*
ID: [your usacogate ID here]
PROG: ride
LANG: C++
*/
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int main() {
    freopen("ride.in", "r", stdin);
    freopen("ride.out", "w", stdout);
    char a[10], b[10];
    int c=1, d=1;
    scanf("%s\n%s", a, b);
    for(int i=0; i<strlen(a); i++) c*=a[i]-'A'+1;
    for(int i=0; i<strlen(b); i++) d*=b[i]-'A'+1;
    c%=47, d%=47;
    if(c==d) printf("GO\n");
    else printf("STAY\n");
    return 0;
}

note: just comment freopen() to test it in ideone

usaco Submitting solutions - Sum the numbers solution

usaco TEXT Submitting solutions - Sum the numbers:

usaco TEXT Submitting solutions - Sum the numbers solution:

/*
ID: [your usacogate ID here]
PROG: test
LANG: C++
*/
#include <iostream>
#include <cstdio>
#include <fstream>
using namespace std;

int main() {
    freopen("test.in", "r", stdin);
    freopen("test.out", "w", stdout);
    int a, b;
    scanf("%d%d", &a, &b);
    printf("%d\n", a+b);
    return 0;
}