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// 求整数1-100的累加值，但要求跳过所有个位为3的数【用continue实现】
var sum = 0;
for (var i = 1; i <= 100; i++) {
if (i % 10 == 3) {//个位为3的数，3/13/23/33/43/53/63/73/83/93
continue;
}
sum += i;
}
console.log(sum);
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