Ultraviolet A 1564-widget Factory (Gaussian element elimination + reverse element)

Link to the question: Ultraviolet A 1564-widget Factory

N kinds of parts, m Work Schedule, Part number from 1 to n, give m work schedule information, X, S, E, indicates that the production of X parts, from week s to week e (it may be multiple weeks), and then the serial number of X parts produced is given. How many days is required for each part to be produced (3 to 10 days guaranteed)

Solution: because it cannot be determined that each calendar has been produced for several days, the corresponding equations listed are linear modulus equations (Module 7 ), therefore, in the process of Gaussian elimination, Division must be converted to multiplication inverse element.

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 305;const int maxd = 7;const char day[maxd][maxd] = {"MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN"};typedef long long ll;typedef ll Mat[maxn][maxn];int N, M;Mat A;int solve (char *s, char *e) {    int p = 0;    while (strcmp(s, day[p])) p++;    int q = p;    while (strcmp(e, day[q])) q = (q + 1) % maxd;    return (q - p + 8) % maxd;}void init () {    int n, x;    char s[maxn], e[maxn];    memset(A, 0, sizeof(A));    for (int i = 0; i < M; i++) {        scanf("%d%s%s", &n, s, e);        for (int j = 0; j < n; j++) {            scanf("%d", &x);            A[i][x-1] = (A[i][x-1] + 1) % maxd;        }        A[i][N] = solve(s, e);    }}ll pow_mod(ll x, int n, ll mod) {    ll ret = 1;    while (n) {        if (n&1)            ret = ret * x % mod;        x = x * x % mod;        n >>= 1;    }    return ret;}ll inv (ll k) {    return pow_mod(k, maxd-2, maxd);}int gauss_elimin (int n, int m) {    int i = 0, j = 0;    while (i < m && j < n) {        int r = i;        for (int k = i; k < m; k++) {            if (A[k][j]) {                r = k;                break;            }        }        if (r != i) {            for (int k = 0; k <= n; k++)                swap(A[i][k], A[r][k]);        }        if (A[i][j] == 0) {            j++;            continue;        }        for (int k = 0; k < m; k++) {            if (A[k][j] == 0 || i == k)                continue;            ll f = A[k][j] * inv(A[i][j]) % maxd;            for (int t = j; t <= n; t++)                A[k][t] = ((A[k][t] - f * A[i][t]) % maxd + maxd) % maxd;        }        i++;    }    for (int k = i; k < m; k++)        if (A[k][n])            return 0;    if (i < n)        return 2;    for (int k = 0; k < n; k++) {        A[k][n] = A[k][n] * inv(A[k][k]) % maxd;        if (A[k][n] < 3)            A[k][n] += maxd;        printf("%lld%c", A[k][n], k == n-1 ? ‘\n‘ : ‘ ‘);    }    return 1;}int main () {    while (~scanf("%d%d", &N, &M) && N) {        init();        int type = gauss_elimin(N, M);        if (type == 0)            printf("Inconsistent data.\n");        else if (type == 2)            printf("Multiple solutions.\n");    }    return 0;}