\u96be\u5ea6&\u63a8\u8350\u6307\u6570\uff1a<\/p>\n
<\/p>\n
\u94fe\u63a5\uff1aCF<\/a> \/ luogu<\/a><\/p>\n Diluc and Kaeya<\/strong><\/p>\n<\/blockquote>\n \u8fea\u5362\u514b\u548c\u51ef\u4e9a<\/strong><\/p>\n The tycoon of a winery empire in Mondstadt, unmatched in every possible way. A thinker in the Knights of Favonius with an exotic appearance.<\/em><\/p>\n<\/blockquote>\n \u8fea\u5362\u514b\uff0c\u8499\u5fb7\u57ce\u7684\u8461\u8404\u9152\u5927\u4ea8\uff0c\u5728\u5404\u79cd\u65b9\u9762\u90fd\u5f88\u65e0\u654c\u3002\u51ef\u4e9a\uff0c\u897f\u98ce\u9a91\u58eb\u56e2\u7684\u4e00\u540d\u601d\u60f3\u5bb6\uff0c\u6709\u7740\u5947\u5f02\u7684\u5916\u8868\u3002<\/em><\/p>\n This time, the brothers are dealing with a strange piece of wood marked with their names. This plank of wood can be represented as a string of $n$ characters. Each character is either a ‘D’ or a ‘K’. You want to make some number of cuts (possibly $0$ ) on this string, partitioning it into several contiguous pieces, each with length at least $1$ . Both brothers act with dignity, so they want to split the wood as evenly as possible. They want to know the maximum number of pieces you can split the wood into such that the ratios of the number of occurrences of ‘D’ to the number of occurrences of ‘K’ in each chunk are the same.<\/p>\n<\/blockquote>\n \u73b0\u5728\uff0c\u8fd9\u5bf9\u5144\u5f1f\u9700\u8981\u5904\u7406\u4e00\u5757\u523b\u7740\u4ed6\u4eec\u540d\u5b57\u7684\u5947\u602a\u7684\u6728\u677f\u3002\u8fd9\u4e2a\u6728\u677f\u53ef\u4ee5\u88ab\u4e00\u4e2a\u957f\u5ea6\u4e3a $ n $ \u7684\u5b57\u7b26\u4e32\u4ee3\u66ff\u3002\u6bcf\u4e00\u4e2a\u5b57\u7b26\u53ea\u80fd\u662f ‘D’ \u6216\u8005\u662f ‘K’ \u3002\u4f60\u60f3\u8981\u5bf9\u8fd9\u4e2a\u5b57\u7b26\u4e32\u8fdb\u884c\u51e0\u6b21\u5206\u5272\uff08\u53ef\u4ee5\u662f $ 0 $ \u6b21\uff09\uff0c\u628a\u5b83\u5206\u6210\u82e5\u5e72\u76f8\u90bb\u7684\u90e8\u5206\uff0c\u6bcf\u4e2a\u90e8\u5206\u7684\u957f\u5ea6\u81f3\u5c11\u662f $ 1 $ \u3002\u8fd9\u5bf9\u5144\u5f1f\u90fd\u5f88\u5c0a\u8d35\uff0c\u6240\u4ee5\u4ed6\u4eec\u60f3\u8981\u5c3d\u53ef\u80fd\u516c\u5e73\u5730\u5206\u5272\u3002\u6240\u4ee5\u5728\u4f60\u5206\u5272\u7684\u6bcf\u5757\u6728\u5934\u91cc ‘K’ \u4e0e ‘D’ \u7684\u6bd4\u4f8b\u5fc5\u987b\u76f8\u540c\u3002\u73b0\u5728\u4ed6\u4eec\u60f3\u77e5\u9053\u80fd\u5c06\u6728\u677f\u6700\u5927\u5206\u5272\u4e3a\u51e0\u5757\u3002<\/p>\n Kaeya, the curious thinker, is interested in the solution for multiple scenarios. He wants to know the answer for every prefix<\/strong> of the given string. Help him to solve this problem!<\/p>\n<\/blockquote>\n \u51ef\u4e9a\uff0c\u8fd9\u4e2a\u597d\u5947\u7684\u601d\u60f3\u5bb6\uff0c\u5bf9\u591a\u91cd\u53ef\u80fd\u53d1\u751f\u60c5\u51b5\u7684\u89e3\u90fd\u611f\u5174\u8da3\u3002\u4ed6\u60f3\u77e5\u9053\u88ab\u7ed9\u4e0e\u7684\u5b57\u7b26\u4e32\u7684\u6bcf\u4e00\u4e2a\u524d\u7f00<\/strong>\u7684\u7b54\u6848\u3002\u5e2e\u52a9\u4ed6\u89e3\u51b3\u8fd9\u4e2a\u95ee\u9898\u5427\uff01<\/p>\n For a string we define a ratio as $a:b$ where ‘D’ appears in it $ a $ times, and ‘K’ appears $ b $ times. Note that $ a $ or $ b $ can equal $ 0 $, but not both. Ratios $ a:b $ and $ c:d $ are considered equal if and only if $ ad=bc $ .<\/p>\n<\/blockquote>\n \u5bf9\u4e8e\u4e00\u4e2a\u5b57\u7b26\u4e32\uff0c\u6211\u4eec\u5b9a\u4e49\u4e00\u4e2a\u6bd4\u4f8b $ a:b $ \u662f ‘D’ \u5728\u5176\u4e2d\u51fa\u73b0 $ a $ \u6b21\uff0c’K’ \u51fa\u73b0 $ b $ \u6b21\u3002\u6ce8\u610f $ a $ \u6216\u8005 $ b $ \u90fd\u53ef\u4ee5\u4e3a\u96f6\u4f46\u4e0d\u80fd\u540c\u65f6\u4e3a\u96f6\u3002\u6bd4\u4f8b $ a:b $ \u548c $ c:d $ \u88ab\u8ba4\u4e3a\u76f8\u540c\u65f6\u5f53\u4e14\u4ec5\u5f53 $ ad=bc $ .<\/p>\n \u7b97\u6cd5\uff1a\u8d2a\u5fc3<\/p>\n \u6570\u636e\u7ed3\u6784\uff1ahash\u8868 \u6216 \u5e73\u8861\u6811\uff08\u8fd9\u91cc\u7684\u5e73\u8861\u6811\u53ef\u4ee5\u7528 \u9898\u76ee\u8ba9\u8f93\u51fa\u5168\u90e8\u524d\u7f00\u7684\u7b54\u6848\uff0c\u6211\u4eec\u5148\u8003\u8651\u5176\u5f31\u5316\u7248\u672c\uff0c\u5373\u53ea\u8003\u8651\u5355\u72ec\u4e00\u4e2a\u5b57\u7b26\u4e32\u7684\u60c5\u51b5\u3002\u663e\u7136\uff0c\u5982\u679c\u6211\u4eec\u5206\u5272\u7684\u5168\u90e8\u5b50\u4e32\u6ee1\u8db3 \u2018D\u2019 \u548c ‘K’ \u7684\u6bd4\u4f8b\u76f8\u540c\uff0c\u90a3\u4e48\u5b83\u4eec\u548c\u539f\u5b57\u7b26\u4e32\u7684 ‘D’ \u4e0e ‘K’ \u7684\u6bd4\u4f8b\u4e5f\u76f8\u540c\u3002\u90a3\u4e48\u6211\u4eec\u53ef\u4ee5\u5f88\u5bb9\u6613\u60f3\u5230\u4e00\u4e2a\u8d2a\u5fc3\u7b97\u6cd5\uff1a\u626b\u63cf\u8fd9\u4e2a\u5b57\u7b26\u4e32\u4e00\u65e6\u9047\u5230\u4e0e\u539f\u5b57\u7b26\u4e32\u6bd4\u4f8b\u76f8\u540c\u7684\u4e00\u4e2a\u524d\u7f00\u5c31\u5206\u5272\uff0c\u7136\u540e\u628a\u8ba1\u6570\u6e05\u96f6\u518d\u7ee7\u7eed\u626b\u6ca1\u626b\u8fc7\u7684\u5b57\u7b26\u4e32\u3002<\/em><\/p>\n \u5982\u4f55\u8bc1\u660e\u8fd9\u4e2a\u8d2a\u5fc3\uff1f<\/p>\n \u53ef\u4ee5\u8003\u8651\u8fd9\u4e2a\u5b57\u7b26\u4e32\u7684\u6700\u77ed\u7684\u548c\u539f\u4e32\u6bd4\u4f8b\u76f8\u540c\u7684\u524d\u7f00\u5982\u679c\u4e0d\u5206\u5272\u80fd\u5426\u66f4\u4f18\uff0c\u7b54\u6848\u663e\u7136\u4e0d\u80fd\u3002\u56e0\u4e3a\u53ea\u6709\u5728\u5b83\u540e\u9762\u518d\u63a5\u4e00\u4e2a\u6bd4\u4f8b\u76f8\u540c\u7684\u5b57\u7b26\u4e32\u624d\u80fd\u6709\u6ee1\u8db3\u9898\u610f\u7684\u5206\u5272\uff0c\u4f46\u8fd9\u6837\u5fc5\u7136\u4f1a\u5bfc\u81f4\u6700\u540e\u7ed3\u679c\u51cf\u5c11\uff0c\u5373\u4e0d\u662f\u6700\u4f18\u3002\u800c\u5728\u8fd9\u4e2a\u6700\u77ed\u524d\u7f00\u540e\u9762\u5206\u5272\u5b8c\u4e4b\u540e\uff0c\u540e\u9762\u7684\u90a3\u4e2a\u5b57\u7b26\u4e32\u53ef\u4ee5\u770b\u505a\u662f\u4e0e\u5f53\u524d\u95ee\u9898\u76f8\u540c\u7684\u5b50\u95ee\u9898\uff08\u9012\u5f52\u601d\u60f3\uff09\uff0c\u56e0\u6b64\u8d2a\u5fc3\u5f97\u8bc1\u3002<\/p>\n \u6211\u4eec\u77e5\u9053\u628a\u4e24\u4e2a\u5206\u5272\u597d\u7684\u5b57\u7b26\u4e32\u91cd\u65b0\u63a5\u5230\u4e00\u8d77\u6bd4\u4f8b\u4e0d\u53d8\uff0c\u518d\u52a0\u4e0a\u6211\u4eec\u521a\u624d\u7684\u8d2a\u5fc3\u7b97\u6cd5\uff0c\u8fd9\u4e2a\u95ee\u9898\u53ef\u4ee5\u8f6c\u5316\u4e3a\u67e5\u8be2\u6709\u591a\u5c11\u4e2a\u524d\u7f00\u7684\u6bd4\u4f8b\u548c\u539f\u4e32\u76f8\u540c\u3002\u90a3\u4e48\u5982\u679c\u7528\u4e8c\u7ef4\u6570\u7ec4\u50a8\u5b58\u663e\u7136\u7a7a\u95f4\u4e0d\u591f\uff0c\u6240\u4ee5\u53ef\u4ee5\u8003\u8651\u7528\u4e00\u4e2a\u6570\u636e\u7ed3\u6784\u6765\u67e5\u8be2\uff0chash\u8868\u548c\u5e73\u8861\u6811\uff08 \u65f6\u95f4\u590d\u6742\u5ea6\uff1a<\/p>\n hash\u8868\u7684\u7248\u672c $ O(n) $\uff0c\u5e73\u8861\u6811\u7684\u7248\u672c $ O(n\\log n) $<\/p>\n \uff08\u4f7f\u7528 \u4ee3\u7801\uff08\u7528hash\u8868\u5b9e\u73b0\u7684\uff09\uff1a<\/p>\n \u96be\u5ea6&\u63a8\u8350\u6307\u6570\uff1a \u601d\u7ef4\u96be\u5ea6\uff1a\u666e\u53ca\/\u63d0\u9ad8- \u5b9e\u73b0\u96be\u5ea6\uff1a\u666e\u53ca\/\u63d0\u9ad8- \u63a8\u8350\u6307\u6570\uff1a5.0 \uff08\u63a8\u8350\u6307\u6570\u6700\u9ad8\u4e3a1 […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[4],"tags":[10],"class_list":["post-36","post","type-post","status-publish","format-standard","hentry","category-4","tag-10"],"_links":{"self":[{"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/posts\/36","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=36"}],"version-history":[{"count":2,"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/posts\/36\/revisions"}],"predecessor-version":[{"id":38,"href":"https:\/\/qwq.cafe\/index.php?rest_route=\/wp\/v2\/posts\/36\/revisions\/38"}],"wp:attachment":[{"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=36"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=36"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/qwq.cafe\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=36"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\u9898\u76ee\u7ffb\u8bd1<\/h2>\n
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\u5410\u5608\uff1a\u624d\u770b\u5b8c\u67d0\u5e94\u6025\u98df\u7269\u7684\u9b3c\u755c\u5c31\u770b\u5230\u8fd9\u4e2a\u9898<\/del><\/p>\n\n
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\u9898\u89e3<\/h2>\n
std::map<\/code> \u4ee3\u66ff\uff09<\/p>\nstd::map<\/code>\uff09\u90fd\u53ef\u4ee5\u80dc\u4efb\u3002<\/p>\nstd::map<\/code> \u7684\u7248\u672c\u53ef\u4ee5\u53bb\u6d1b\u8c37\u770b\u5176\u4ed6\u4eba\u7684\u9898\u89e3\uff0c\u56e0\u4e3a\u6709\u522b\u4eba\u5199\u4e86\u6211\u5c31\u6362\u4e00\u4e2a\uff09<\/p>\n#include<stdio.h>\n#include<string.h>\n#define MAXN 500003\nstruct hash_node {\n int x,y;\n int sum;\n int next;\n}h[MAXN];\nint tot;\nint head[MAXN];\nint hashcode(int x,int y)\n{\n return (x*1331+y)%MAXN;\n}\nint add_hash(int x,int y)\n{\n int code=hashcode(x,y),p;\n for(p=head[code];p;p=h[p].next)\n {\n if(h[p].x==x&&h[p].y==y)\n {\n h[p].sum++;\n return h[p].sum;\n }\n }\n h[++tot]=(hash_node){x,y,1,head[code]};\n head[code]=tot;\n return 1;\n}\nint gcd(int x,int y)\n{\n int r=x%y;\n while(r)\n {\n x=y;\n y=r;\n r=x%y;\n }\n return y;\n}\nint main()\n{\n int T,i,n,dsum,ksum,x,y,g;\n char c;\n scanf("%d",&T);\n while(T--)\n {\n tot=dsum=ksum=0;\n memset(h,0,sizeof(h));\n memset(head,0,sizeof(head));\n\n scanf("%d",&n);\n for(i=0;i<n;i++)\n {\n scanf(" %c",&c);\n if(c=='D') dsum++;\n else ksum++;\n if(dsum==0) x=0,y=1;\n else if(ksum==0) x=1,y=0;\n else\n {\n g=gcd(dsum,ksum);\n x=dsum\/g,y=ksum\/g;\n }\n g=add_hash(x,y);\n printf("%d ",g);\n }\n printf("\\n");\n }\n return 0;\n}<\/code><\/pre>\n","protected":false},"excerpt":{"rendered":"