分析 这道题数据范围较小,看题目问题,可确定是一个dfs,每次要么走'-',要么走'+',时间复杂度大概为O( ),可以小剪枝一下,如果当前加上所有的数如果都小于最后一个数,就不用搜了,或者是减去后面所有数都大于最后的数,也不用再搜下去了 代码 //#pragma GCC optimize(3,"inline","Ofast","fast-math","no-stack-protector","unroll-loops") //#pragma GCC target("sse","sse2","sse3","sse4","avx","avx2","popcnt") /* (写点什么吧...)...