cf：D.BlackandWhiteStripe【连续k个中最少的个数+滑动窗口】

分析

每次l和r前进一个
分析头和尾的元素&＃xff0c;来得到当前窗口中最少的个数即可

Ac code

for t in range(int(input())):#print(str(t) &＃43; &＃39;:&＃39;)n, k &＃61; list(map(int, input().split()))s &＃61; input()cntW &＃61; 0ans &＃61; 0xfffffffffor i in range(k):if s[i] &＃61;&＃61; &＃39;W&＃39;:cntW &＃43;&＃61; 1ans &＃61; min(ans, cntW)for i in range(k, n):if s[i] &＃61;&＃61; &＃39;W&＃39; and s[i - k] &＃61;&＃61; &＃39;W&＃39;:passelif s[i] &＃61;&＃61; &＃39;W&＃39; and s[i - k] &＃61;&＃61; &＃39;B&＃39;:cntW &＃43;&＃61; 1elif s[i] &＃61;&＃61; &＃39;B&＃39; and s[i - k] &＃61;&＃61; &＃39;W&＃39;:cntW -&＃61; 1elif s[i] &＃61;&＃61; &＃39;B&＃39; and s[i - k] &＃61;&＃61; &＃39;B&＃39;:passans &＃61; min(ans, cntW)print(ans)


总结

思维转换 &＃43; 滑动窗口