poj 2947 Widget Factory（高斯消元）

根据样例列出来方程，解出来结果是1，3，实际结果是8，3。。懵逼了，一周最多七天，怎么会有8。搜了下题解才知道这中间可能隔着很多周，那这就是用高斯消元解线性同余方程组了，而且方程很好列，把kuangbin大佬板子拿来就是了。（板子就是这道题）
具体意思看这里：http://blog.csdn.net/qingshui23/article/details/52511894

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;const int mod = 7;const int MAXN = 400;int a[MAXN][MAXN];int x[MAXN];int gcd(int a, int b){    if(b == 0) return a;    return gcd(b,a%b);}int lcm(int a, int b){    return a/gcd(a,b)*b;}long long inv(long long a, long long m){    if(a == 1) return 1;    return inv(m%a,a)*(m-m/a)%m;}int change(char* s){    if(strcmp(s,"MON") == 0) return 1;    else if(strcmp(s,"TUE")==0) return 2;    else if(strcmp(s,"WED")==0) return 3;    else if(strcmp(s,"THU")==0) return 4;    else if(strcmp(s,"FRI")==0) return 5;    else if(strcmp(s,"SAT")==0) return 6;    else return 7;}//equ行，var列int Gauss(int equ, int var){    int maxR,col,k;    for(k = 0, col = 0; k < equ && col < var; ++k,++col)    {        maxR = k;        for(int i = k+1; i < equ; ++i)            if(abs(a[i][col]) > abs(a[maxR][col]))                maxR = i;        if(a[maxR][col] == 0)        {            k--;            continue;        }        if(maxR != k)            for(int j = col; j < var+1; ++j)                swap(a[k][j],a[maxR][j]);        for(int i = k+1; i < equ; ++i)        {            if(a[i][col] != 0)            {                int LCM = lcm(abs(a[i][col]),abs(a[k][col]));                int ta = LCM/abs(a[i][col]);                int tb = LCM/abs(a[k][col]);                if(a[i][col]*a[k][col] < 0) tb = -tb;                for(int j = col; j < var+1; ++j)                    a[i][j] = ((a[i][j]*ta-a[k][j]*tb)%mod+mod)%mod;            }        }    }    for(int i = k; i < equ; ++i)        if(a[i][col] != 0)            return -1;    if(k < var) return var-k;    for(int i = var-1; i >= 0; --i)    {        int temp = a[i][var];        for(int j = i+1; j < var; ++j)        {            if(a[i][j] != 0)            {                temp -= a[i][j]*x[j];                temp = (temp%mod+mod)%mod;            }        }        x[i] = (temp*inv(a[i][i],mod))%mod;    }    return 0;}int main(){    int n,m;    while(scanf("%d %d",&n,&m) && n+m)    {        memset(a,0,sizeof(a));        char str1[10],str2[10];        int k;        for(int i = 0; i < m; ++i)        {            scanf("%d %s %s",&k,str1,str2);            a[i][n] = ((change(str2)-change(str1)+1)%mod+mod)%mod;            int t;            while(k--)            {                scanf("%d",&t);                t--;                a[i][t]++;                a[i][t] %= mod;            }        }        int res = Gauss(m,n);        if(res == 0)        {            for(int i = 0; i < n; ++i)                if(x[i] <= 2)                    x[i] += 7;            for(int i = 0; i < n-1; ++i)                printf("%d ",x[i]);            printf("%d\n",x[n-1]);        }        else if(res == -1) printf("Inconsistent data.\n");        else printf("Multiple solutions.\n");    }    return 0;}