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2 forfor (i = 0; i != n; i++) { k = 0; for (j = 0; (i + j) != n; j++) { k += b[i + j]; if (k > 10000) break; c[k]++; } n是素数个数 b存素数[2, 3, 5, 7...] c存表示方法数,初始为0 Followed by: Post your reply here: |
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