London Symphony Orchestra & Nikolaj Znaider
Nikolaj Znaider - Mozart: Violin Concertos No. 1, 2 & 3 (24-96)
2018
Wolfgang Amadeus Mozart
01 Violin Concerto No. 1 in B flat major, K. 207: I. Allegro moderato
02 Violin Concerto No. 1 in B flat major, K. 207: II. Adagio
03 Violin Concerto No. 1 in B flat major, K. 207: III. Presto
04 Violin Concerto No. 2 in D major, K. 211: I. Allegro moderato
05 Violin Concerto No. 2 in D major, K. 211: II. Andante
06 Violin Concerto No. 2 in D major, K. 211: III. Rondeau. Allegro
07 Violin Concerto No. 3 in G major, K. 216 'Strassburg': I. Allegro
08 Violin Concerto No. 3 in G major, K. 216 'Strassburg': II. Adagio
09 Violin Concerto No. 3 in G major, K. 216 'Strassburg': III. Rondeau. Allegro
Nikolaj Znaider - Mozart: Violin Concertos No. 1, 2 & 3 (24-96)
2018
Wolfgang Amadeus Mozart
01 Violin Concerto No. 1 in B flat major, K. 207: I. Allegro moderato
02 Violin Concerto No. 1 in B flat major, K. 207: II. Adagio
03 Violin Concerto No. 1 in B flat major, K. 207: III. Presto
04 Violin Concerto No. 2 in D major, K. 211: I. Allegro moderato
05 Violin Concerto No. 2 in D major, K. 211: II. Andante
06 Violin Concerto No. 2 in D major, K. 211: III. Rondeau. Allegro
07 Violin Concerto No. 3 in G major, K. 216 'Strassburg': I. Allegro
08 Violin Concerto No. 3 in G major, K. 216 'Strassburg': II. Adagio
09 Violin Concerto No. 3 in G major, K. 216 'Strassburg': III. Rondeau. Allegro
【走出精神内耗,为生活“节能”】
#京彩首图# #主题# #周一综合症# #荐书# 《山月记》
馆藏地:首图B座三层文学图书
索书号:I313.45 /3150
“我深怕自己本非美玉,故而不敢加以刻苦琢磨,却又半信自己是块美玉,故又不肯庸庸碌碌,与瓦砾为伍。”
这种心境你是否有过呢?
有时觉得天生我材必有用,有时又懦弱自卑,害怕努力之后更加暴露出自己才华不足。
在古典与现代之间,读懂《山月记》,就读懂自己内心的焦虑。
#京彩首图# #主题# #周一综合症# #荐书# 《山月记》
馆藏地:首图B座三层文学图书
索书号:I313.45 /3150
“我深怕自己本非美玉,故而不敢加以刻苦琢磨,却又半信自己是块美玉,故又不肯庸庸碌碌,与瓦砾为伍。”
这种心境你是否有过呢?
有时觉得天生我材必有用,有时又懦弱自卑,害怕努力之后更加暴露出自己才华不足。
在古典与现代之间,读懂《山月记》,就读懂自己内心的焦虑。
#小全不努力怎么行#
题目:给两个整数数组 nums1 和 nums2 ,返回 两个数组中 公共的 、长度最长的子数组的长度 。
解:动态规划。
dp[i][j] :以下标i - 1为结尾的A,和以下标j - 1为结尾的B,最⻓重复⼦数组⻓度为dp[i][j]。dp[0][0]可以看作下标-1的数组,这样才能记录数组下标为0的数。
当nums1[i-1]==nums2[j-1]时,dp[i][j] = dp[i - 1][j - 1] + 1;并用result记录最大值。
题目:给两个整数数组 nums1 和 nums2 ,返回 两个数组中 公共的 、长度最长的子数组的长度 。
解:动态规划。
dp[i][j] :以下标i - 1为结尾的A,和以下标j - 1为结尾的B,最⻓重复⼦数组⻓度为dp[i][j]。dp[0][0]可以看作下标-1的数组,这样才能记录数组下标为0的数。
当nums1[i-1]==nums2[j-1]时,dp[i][j] = dp[i - 1][j - 1] + 1;并用result记录最大值。
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