如何用C+编译出来一个9*9的数
#include typedef struct{ int line; int row; int num;}Node;int main(){ /* int a[9][9]={ {4,0,3,6,0,0,0,0,0}, {0,0,0,0,0,1,0,2,4}, {0,1,0,0,4,0,5,0,0}, {0,0,0,9,0,4,0,6,0}, {3,0,2,0,0,0,4,0,9}, {0,7,4,1,0,3,0,0,0}, {0,0,1,0,9,0,0,4,0}, {2,4,0,3,0,0,0,0,0}, {0,0,0,4,0,8,2,0,7}}; */ int a[9]...全部
#include typedef struct{ int line; int row; int num;}Node;int main(){ /* int a[9][9]={ {4,0,3,6,0,0,0,0,0}, {0,0,0,0,0,1,0,2,4}, {0,1,0,0,4,0,5,0,0}, {0,0,0,9,0,4,0,6,0}, {3,0,2,0,0,0,4,0,9}, {0,7,4,1,0,3,0,0,0}, {0,0,1,0,9,0,0,4,0}, {2,4,0,3,0,0,0,0,0}, {0,0,0,4,0,8,2,0,7}}; */ int a[9][9]={ {0,0,0,8,0,0,0,6,0}, {8,7,0,0,0,0,0,0,0}, {2,9,0,0,4,1,0,0,5}, {0,0,5,7,0,0,0,0,9}, {0,2,0,0,0,0,0,1,0}, {9,0,0,0,0,4,3,0,0}, {7,0,0,6,1,0,0,9,8}, {0,0,0,0,0,0,0,5,2}, {0,6,0,0,0,9,0,0,0}}; /* int a[9][9]={ {0,2,0,0,6,0,0,0,0}, {0,9,0,4,0,5,1,3,0}, {0,0,8,7,0,0,0,0,5}, {6,0,0,3,0,0,4,0,0}, {0,0,0,9,0,6,0,0,0}, {0,0,7,0,0,1,0,0,3}, {4,0,0,0,0,7,3,0,0}, {0,8,5,2,0,4,0,7,0}, {0,0,0,0,9,0,0,1,0}}; */ /* int a[9][9]={ {0,0,3,0,2,0,0,0,6}, {0,0,2,0,9,0,0,0,4}, {7,0,0,8,0,0,2,0,3}, {0,8,0,0,7,0,5,0,0}, {0,7,0,1,0,6,0,3,0}, {0,0,0,2,0,0,0,9,0}, {4,0,6,0,0,8,0,0,5}, {6,0,0,0,4,0,3,0,0}, {9,0,0,0,1,0,7,0,0}}; */ int i,j,n,en,flag,y,k=0,x,qu,p,q; Node b[70]; for(i=0;i9,则k-1,否则k+1*/ for(k=0;k9) { a[i][j]=b[k]。
num=0; k--; if(k<0) { printf("error!\r\n"); return -1; } } else k++; } for(i=0;i<9;i++) { for(j=0;j<9;j++) { printf("%d",a[i][j]); } printf("\r\n"); } return 1;}
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