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通过循环的询问可以找到candidate,每次询问排除两者之一最后检查一下这个candidate是否valid int candidate = 0; for(int i = 1; i < n; i++){ if(knows(candidate, i)) { candidate = i; } } for(int i = 0; i < n; i++){ if(i != candidate && (knows(candidate, i) || !knows(i, candidate))) { return -1; } } return candidate;
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