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char c[10][10]={
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\"\",
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\"\",
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\"ABC\",
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\"DEF\",
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\"GHI\",
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\"JKL\",
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\"MNO\",
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\"PQRS\",
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\"TUV\",
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\"WXYZ\"
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}
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int total[10]={ 0,0,3,3,3,3,3,4,3,4};
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const int TelLength=3;
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int number[TelLength];
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int answer[TelLength];
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//方法一
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while(true)
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{
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for(i=0; i<TelLength; i++)
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printf(\"%c\", c[number[i]][answer[i]]);
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printf(\"\\n\");
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int k=n-1;
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while(k>=0)
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{
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if(answer[k]<total[number[k]]-1)
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{
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answer[k]++;
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break;
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}
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else
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{
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answer[k]=0; k--;
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}
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}
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if(k<0)
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break;
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}
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//方法二
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void Recursion(int *number, int *answer, int index, int n)
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{
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if(index==n)
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{
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for(int i=0; i<n; i++)
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{
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printf(\"%c\", c[number[i]][answer[i]]);
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}
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printf(\"\\n\");
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return;
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}
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for( answer[index]=0; answer[index]<total[number[index]]; answer[index]++)
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{
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RecursiveSearch(number, answer, index+1, n);
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}
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来自 “ ITPUB博客 ” ,链接:http://blog.itpub.net/29012686/viewspace-1145084/,如需转载,请注明出处,否则将追究法律责任。
转载于:http://blog.itpub.net/29012686/viewspace-1145084/
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