Cod sursa(job #1994420)

Utilizator giotoPopescu Ioan gioto Data 24 iunie 2017 22:09:21
Problema Perb Scor 70
Compilator cpp Status done
Runda Arhiva de probleme Marime 2.56 kb
#include <algorithm>
#include <fstream>
#include <cstdio>
using namespace std;

int n, m, best[605][605], dif[605][128];
char s[605];
class InParser {
private:
    FILE *fin;
    char *buff;
    int sp;

    char read_ch() {
        ++sp;
        if (sp == 4096) {
            sp = 0;
            fread(buff, 1, 4096, fin);
        }
        return buff[sp];
    }

public:
    InParser(const char* nume) {
        fin = fopen(nume, "r");
        buff = new char[4096]();
        sp = 4095;
    }

    InParser& operator >> (int &n) {
        char c;
        while (!isdigit(c = read_ch()) && c != '-');
        int sgn = 1;
        if (c == '-') {
            n = 0;
            sgn = -1;
        } else {
            n = c - '0';
        }
        while (isdigit(c = read_ch())) {
            n = 10 * n + c - '0';
        }
        n *= sgn;
        return *this;
    }

    InParser& operator >> (long long &n) {
        char c;
        n = 0;
        while (!isdigit(c = read_ch()) && c != '-');
        long long sgn = 1;
        if (c == '-') {
            n = 0;
            sgn = -1;
        } else {
            n = c - '0';
        }
        while (isdigit(c = read_ch())) {
            n = 10 * n + c - '0';
        }
        n *= sgn;
        return *this;
    }
};
int main()
{
    freopen("perb.in", "r", stdin);
    scanf("%d%d", &n, &m);
    scanf("%s", s + 1);
    fclose(stdin);
    for(int i = 1; i <= n ; ++i)
        for(int j = i; j <= n ; ++j)
            best[i][j] = j - i + 1;
    for(int i = 1; i < n ; ++i){
        for(int d = 1; d < n - 1 ; ++d){
            for(int j = 0; j < d ; ++j) dif[j]['A'] = dif[j]['C'] = dif[j]['G'] = dif[j]['T'] = 0;
            int aux = 0, rest = 0;
            for(int j = i; j <= n ; ++j){
                ++rest;
                if(rest == d) rest = 0;
                aux = aux - (dif[rest]['A'] + dif[rest]['C'] + dif[rest]['G'] + dif[rest]['T'] - max(max(dif[rest]['A'], dif[rest]['C']), max(dif[rest]['G'], dif[rest]['T'])));
                ++dif[rest][s[j]];
                aux = aux + (dif[rest]['A'] + dif[rest]['C'] + dif[rest]['G'] + dif[rest]['T'] - max(max(dif[rest]['A'], dif[rest]['C']), max(dif[rest]['G'], dif[rest]['T'])));
                if(rest == 0 && j - i + 1 > d) best[i][j] = min(best[i][j], aux);
            }
        }
    }
    InParser fin("perb.in");
    ofstream fout("perb.out");
    int l, r;
    fin >> l >> r;
    for(int i = 1; i <= m ; ++i){
        fin >> l >> r;
        fout <<  best[l][r] << "\n";
    }
    return 0;
}