poj 2947

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题目描述：

就是求模7的方程组.然后结果在3到9之间.n~300
题解：

高斯消元.用公倍数和除以逆元来解.
重点：

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cmath>#include <ctype.h>#include <limits.h>#include <cstdlib>#include <algorithm>#include <vector>#include <queue>#include <map>#include <stack>#include <set>#include <bitset>#define CLR(a) memset(a, 0, sizeof(a))#define REP(i, a, b) for(int i = a;i < b;i++)#define REP_D(i, a, b) for(int i = a;i <= b;i++)typedef long long ll;using namespace std;const int maxn = 300+100;const int MOD=7;int m,n;int a[maxn][maxn], x[maxn];int equ, var;inline int gcd(int a,int b){    while(b != 0)    {        int t = b;        b = a%b;        a = t;    }    return a;}inline 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;    int t = pow(a, m-2) + 1e-9;    return t%MOD;}int gauss(){    int k, col;    for(k=0,col=0; k<equ&&col<var; k++,col++)    {        int max_r = k;        for(int i = k+1; i < equ; i++)            if(abs(a[i][col]) > abs(a[max_r][col]))                max_r = i;        if(a[max_r][col] == 0)        {            k--;            continue;        }        if(max_r != k)            for(int j = col; j < var+1; j++)                swap(a[k][j],a[max_r][j]);        for(int i=k+1; i<equ; i++)        {            if(a[i][col]!=0)            {                int d = lcm(abs(a[i][col]), abs(a[k][col]) );                int d_k = d/a[k][col], d_i = d/a[i][col];                if(a[i][col]*a[k][col]<0)                    d_i=-d_i;                for(int j=col; j<=var; j++)                {                    int lft = ((a[i][j]*d_i-a[k][j]*d_k)%MOD+MOD)%MOD;                    a[i][j]=lft;                }            }        }    }    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++)        {            temp = ((temp-(a[i][j]*x[j])%MOD)%MOD+MOD)%MOD;        }        x[i] = (temp*inv(a[i][i],MOD))%MOD;    }    return 0;}void solve(){    equ=m;    var=n;    int t=gauss();    if(t==-1)    {        printf("Inconsistent data.\n");    }    else if(t>0)    {        printf("Multiple solutions.\n");    }    else    {        for(int i=0; i<var; i++)        {            if(x[i]<=2)                x[i] += 7;            printf("%d%c", x[i], (i==var-1 ? '\n' : ' '));        }    }}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;}int main(){    freopen("5Ein.txt", "r", stdin);    //freopen("5Eout.txt", "w", stdout);    while(scanf("%d%d",&n,&m) == 2)    {        if(n == 0 && m == 0)break;        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;            }        }        solve();    }    return 0;}
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