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居然ac了#include<iostream> using namespace std; long long int a[4][5]; int main(){ int n,i,j; long long int det1,det2,det3,det; cin>>n; double x; while(n--){ for(i=1;i<=3;i++) for(j=1;j<=4;j++) scanf("%lld",&a[i][j]); det=a[1][1]*(a[2][2]*a[3][3]-a[2][3]*a[3][2])-a[1][2]*(a[2][1]*a[3][3]-a[2][3]*a[3][1])+a[1][3]*(a[2][1]*a[3][2]-a[2][2]*a[3][1]); det1=a[1][4]*(a[2][2]*a[3][3]-a[2][3]*a[3][2])-a[1][2]*(a[2][4]*a[3][3]-a[2][3]*a[3][4])+a[1][3]*(a[2][4]*a[3][2]-a[2][2]*a[3][4]); det2=a[1][1]*(a[2][4]*a[3][3]-a[2][3]*a[3][4])-a[1][4]*(a[2][1]*a[3][3]-a[2][3]*a[3][1])+a[1][3]*(a[2][1]*a[3][4]-a[2][4]*a[3][1]); det3=a[1][1]*(a[2][2]*a[3][4]-a[2][4]*a[3][2])-a[1][2]*(a[2][1]*a[3][4]-a[2][4]*a[3][1])+a[1][4]*(a[2][1]*a[3][2]-a[2][2]*a[3][1]); printf("%lld %lld %lld %lld\n",det1,det2,det3,det); if(det==0) printf("No unique solution\n"); else {printf("Unique solution: "); x=(double)det1/(double)det; if(x>-0.0005&&x<0.0005) x=0; printf("%.3lf ",x); x=(double)det2/(double)det; if(x>-0.0005&&x<0.0005) x=0; printf("%.3lf ",x); x=(double)det3/(double)det; if(x>-0.0005&&x<0.0005) x=0; printf("%.3lf\n",x); } printf("\n"); } return 1; } Followed by: Post your reply here: |
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