string(14) "127.0.0.1:9200"
URL: http://127.0.0.1:9200/likecs_art_db/_search<br>REQUEST:<pre>Array
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            [pre_tags] => #em#
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</pre>RESPONSE:<pre>string(8220) "{"took":20,"timed_out":false,"_shards":{"total":1,"successful":1,"skipped":0,"failed":0},"hits":{"total":{"value":10000,"relation":"gte"},"max_score":83.53072,"hits":[{"_index":"likecs_art_db","_type":"_doc","_id":"72596","_score":83.53072,"_source":{"id":"72596","text":"2019\u725b\u5ba2\u6691\u671f\u591a\u6821\u8bad\u7ec3\u8425\uff08\u7b2c\u4e8c\u573a\uff09J-Subarray","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        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   color: [\r\n        ","username":"wizarderror","tagsname":"","tagsid":"[]","catesname":"","catesid":"[]","createtime":"1564324531"},"highlight":{"text":["#em#2019#/em##em#牛#/em##em#客#/em##em#暑#/em##em#期#/em##em#多#/em##em#校#/em##em#训#/em##em#练#/em##em#营#/em#（#em#第#/em#四#em#场#/em#）#em#J#/em#-free"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"70626","_score":59.51892,"_source":{"id":"70626","text":"2019\u725b\u5ba2\u6691\u671f\u591a\u6821\u8bad\u7ec3\u8425\uff08\u7b2c\u5341\u573a\uff09J - Wood Processing 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(斜率优化DP)"]}},{"_index":"likecs_art_db","_type":"_doc","_id":"203380951","_score":56.808765,"_source":{"id":"203380951","text":"\u725b\u5ba2\u7f51\u6691\u671fACM\u591a\u6821\u8bad\u7ec3\u8425\uff08\u7b2c\u4e00\u573a\uff09A","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        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</pre>string(14) "127.0.0.1:9200"
URL: http://127.0.0.1:9200/likecs_art_db/_search<br>REQUEST:<pre>Array
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                            [query] => 2019牛客暑期多校训练营（第二场）J-Subarray
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            [pre_tags] => #em#
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    [size] => 8
    [from] => 8
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</pre>RESPONSE:<pre>string(8450) "{"took":21,"timed_out":false,"_shards":{"total":1,"successful":1,"skipped":0,"failed":0},"hits":{"total":{"value":10000,"relation":"gte"},"max_score":83.53072,"hits":[{"_index":"likecs_art_db","_type":"_doc","_id":"235298","_score":54.984158,"_source":{"id":"235298","text":"2021\u725b\u5ba2\u6691\u671f\u591a\u6821\u8bad\u7ec3\u84258","intro":"\u76ee\u5f55\n\nECharts\n\u5f02\u6b65\u52a0\u8f7d\n\n\n\nECharts\r\n\u6570\u636e\u53ef\u89c6\u5316\u5728\u8fc7\u53bb\u51e0\u5e74\u4e2d\u53d6\u5f97\u4e86\u5de8\u5927\u8fdb\u5c55\u3002\u5f00\u53d1\u4eba\u5458\u5bf9\u53ef\u89c6\u5316\u4ea7\u54c1\u7684\u671f\u671b\u4e0d\u518d\u662f\u7b80\u5355\u7684\u56fe\u8868\u521b\u5efa\u5de5\u5177\uff0c\u800c\u662f\u5728\u4ea4\u4e92\u3001\u6027\u80fd\u3001\u6570\u636e\u5904\u7406\u7b49\u65b9\u9762\u6709\u66f4\u9ad8\u7684\u8981\u6c42\u3002\r\nchart.setOption({\r\n    color: [\r\n        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</pre>string(14) "127.0.0.1:9200"
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</pre>string(14) "127.0.0.1:9200"
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                        <h1 class="page-title"><a href="/show-72596.html" title="2019牛客暑期多校训练营（第二场）J-Subarray">2019牛客暑期多校训练营（第二场）J-Subarray</a></h1>
            
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<div id="likecs_post_body" class="blogpost-body ">
    <p><a style="cursor: pointer;" onclick="window.open('https://ac.nowcoder.com/acm/contest/882/J','','');return false;" href="https://ac.nowcoder.com/acm/contest/882/J" target="_blank" rel="nofollow"><span style="font-size: 18px;">&gt;传送门&lt;</span></a></p>
<p><strong><span style="font-size: 18pt;">前言</span></strong></p>
<hr>
<p><span style="font-size: 18px;">这题我前前后后看了三遍，每次都是把网上相关的博客和通过代码认真看了再思考，然并卵，最后终于第三遍也就是现在终于看懂了，其实懂了之后发现其实没有那么难，但是的的确确需要思维。(博客分析那块写的啰里吧嗦又改了很多废话)</span></p>
<p><strong><span style="font-size: 18pt;">题意</span></strong></p>
<hr>
<p><span style="font-size: 18px;">在一个长度为$10^{9}$的序列上，保证只有$n(n&lt;10^{6})$个区间等于$1$，且$1$的个数小于10^{7}，其他位置全部为$-1$，求区间和$&gt;0$的区间数量</span></p>
<p><strong><span style="font-size: 18pt;">分析</span></strong></p>
<hr>
<p><span style="font-size: 18px;">题目意思很简单，对不对~ 那接下来我们就思考下该怎么做这题。</span></p>
<p><span style="font-size: 18px;">考虑做前缀和，问题就转化成$sum[i]-sum[j] &gt; 0$的对数</span></p>
<p><span style="font-size: 18px;">举个栗子，比如序列$\left \{ -1,-1,-1,1,1,1,1 \right \}$，对应的前缀和为<span style="background-color: #ffffff; color: #000000;">$\left \{ -1,-2,-3,-2,-1,0,1 \right \}$</span></span></p>
<p><span style="font-size: 18px;">那么$-1&gt;-2$，中间的$\left \{ -1,-1,1,1,1 \right \}$区间和$&gt;0$</span></p>
<p><span style="font-size: 18px;">由于数据范围较大不可能对整个数组前缀和进行处理，我们注意到<strong>能被1覆盖的区间长度最多有$3e7$， 因为区间向前，向后最多可以覆盖$1e7$，所以加起来就是$3e7$，这是本题求解的关键。</strong></span></p>
<p><span style="font-size: 18px;">所以我们可以处理一下区间，让它尽可能的延伸而又不至于小于$0$，之后我们只需要计算在当前点有多少前面点的前缀和小于它就行了。这样我们就可以用树状数组来求，可范围太大了我们只能用别的方法(不甘心的我用树状数组试了下果然超时)</span></p>
<p><span style="font-size: 14pt; background-color: #ccffff;"><strong>Code</strong></span></p>
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<pre>#include&lt;bits/stdc++.h&gt;<span style="color: #000000;">
typedef </span><span style="color: #0000ff;">long</span> <span style="color: #0000ff;">long</span><span style="color: #000000;"> ll;
</span><span style="color: #0000ff;">using</span> <span style="color: #0000ff;">namespace</span><span style="color: #000000;"> std;
</span><span style="color: #0000ff;">const</span> <span style="color: #0000ff;">int</span> maxn = 1e7+<span style="color: #800080;">10</span><span style="color: #000000;">; 
</span><span style="color: #0000ff;">int</span><span style="color: #000000;"> l[maxn], r[maxn], n;
</span><span style="color: #0000ff;">int</span><span style="color: #000000;"> L[maxn],R[maxn];
</span><span style="color: #0000ff;">int</span> num[maxn*<span style="color: #800080;">3</span><span style="color: #000000;">]; 
</span><span style="color: #0000ff;">int</span><span style="color: #000000;"> main()
{
    scanf(</span><span style="color: #800000;">"</span><span style="color: #800000;">%d</span><span style="color: #800000;">"</span>, &amp;<span style="color: #000000;">n);
    </span><span style="color: #0000ff;">for</span>(<span style="color: #0000ff;">int</span> i = <span style="color: #800080;">1</span>; i &lt;= n; i++) scanf(<span style="color: #800000;">"</span><span style="color: #800000;">%d%d</span><span style="color: #800000;">"</span>, &amp;l[i], &amp;<span style="color: #000000;">r[i]);
    r[</span><span style="color: #800080;">0</span>] = R[<span style="color: #800080;">0</span>] = -<span style="color: #800080;">1</span>, l[n+<span style="color: #800080;">1</span>] =<span style="color: #000000;"> 1e9;
    </span><span style="color: #0000ff;">int</span> len=<span style="color: #800080;">0</span><span style="color: #000000;">;
    </span><span style="color: #0000ff;">for</span>(<span style="color: #0000ff;">int</span> i = <span style="color: #800080;">1</span>; i &lt;= n; i++<span style="color: #000000;">){
        len </span>+= r[i]-l[i]+<span style="color: #800080;">1</span><span style="color: #000000;">;
        R[i] </span>= min(r[i]+len,l[i+<span style="color: #800080;">1</span>]-<span style="color: #800080;">1</span><span style="color: #000000;">);
        len </span>-= l[i+<span style="color: #800080;">1</span>]-r[i]-<span style="color: #800080;">1</span><span style="color: #000000;">;
        </span><span style="color: #0000ff;">if</span>(len&lt;<span style="color: #800080;">0</span>) len=<span style="color: #800080;">0</span><span style="color: #000000;">;
    }
    len</span>=<span style="color: #800080;">0</span><span style="color: #000000;">;
    </span><span style="color: #0000ff;">for</span>(<span style="color: #0000ff;">int</span> i = n; i &gt; <span style="color: #800080;">0</span>; i--<span style="color: #000000;">){
        len </span>+= r[i]-l[i]+<span style="color: #800080;">1</span><span style="color: #000000;">;
        L[i] </span>= max(l[i]-len,r[i-<span style="color: #800080;">1</span>]+<span style="color: #800080;">1</span><span style="color: #000000;">);
        len </span>-= l[i]-r[i-<span style="color: #800080;">1</span>]-<span style="color: #800080;">1</span><span style="color: #000000;">;
        </span><span style="color: #0000ff;">if</span>(len&lt;<span style="color: #800080;">0</span>) len=<span style="color: #800080;">0</span><span style="color: #000000;">;
    }
    
    </span><span style="color: #0000ff;">int</span> now = 1e7+<span style="color: #800080;">10</span><span style="color: #000000;">;
    num[now] </span>= <span style="color: #800080;">1</span><span style="color: #000000;">;
    ll sum </span>= <span style="color: #800080;">0</span>, ans = <span style="color: #800080;">0</span><span style="color: #000000;">;
    </span><span style="color: #0000ff;">for</span>(<span style="color: #0000ff;">int</span> i = <span style="color: #800080;">1</span>; i &lt;= n; i++<span style="color: #000000;">){
        </span><span style="color: #0000ff;">for</span>(<span style="color: #0000ff;">int</span> j = max(L[i],R[i-<span style="color: #800080;">1</span>]+<span style="color: #800080;">1</span>); j &lt;= R[i]; j++<span style="color: #000000;">){
            </span><span style="color: #0000ff;">if</span>(j&gt;=l[i]&amp;&amp;j&lt;=<span style="color: #000000;">r[i]){
                sum </span>+=<span style="color: #000000;"> num[now];
                num[</span>++now]++<span style="color: #000000;">;
            }</span><span style="color: #0000ff;">else</span><span style="color: #000000;">{
                sum </span>-= num[--<span style="color: #000000;">now];
                num[now]</span>++<span style="color: #000000;">;
            }
            ans </span>+=<span style="color: #000000;"> sum;
        }
    }
    printf(</span><span style="color: #800000;">"</span><span style="color: #800000;">%lld</span><span style="color: #800000;">"</span><span style="color: #000000;">,ans);
    </span><span style="color: #0000ff;">return</span> <span style="color: #800080;">0</span><span style="color: #000000;">;
}

</span><span style="color: #008000;">/*</span><span style="color: #008000;">
now记录的是当前的前缀和，sum记录前缀和小于now的数量，num数组记录的所有前缀和对应的数目。
假如j对应的数字是1，num[++now]显然应该加1，由于此时前缀和比上次要大，那么执行这一步操作前我们应该先把sum加上num[now]；
假如j对应的数字是-1，num[--now]也当然加1，这时候前缀和比上次要小，执行完这步后sum应当减去此时和now相等的前缀和数量就得到比now小的前缀和数量。
</span><span style="color: #008000;">*/</span></pre>
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</div>
<p> <span style="text-decoration: underline;"><em><span style="font-size: 18px;">可能有的人会跟我有一样的疑惑，一但中间出现“断层”怎么办？也即是中间有非常多的$-1$导致相邻两端不能相连。你仔细想想，其实对于我们这种做法没有任何影响，<strong>我们只处理对答案有贡献的区间的点</strong>。你可能会说如果”断层“的话前缀和应该重新计算，但是实际上我们是在判断$sum[i]-sum[j] &gt; 0$即$sum[i]&gt;sum[j]$，从不从$0$开始算我们并不关心，我们只在乎他们的相对大小，这里一定要好好想一想！！！</span></em></span></p>
<p> </p>
<hr>
<p><span style="font-size: 16px;">参考资料：</span></p>
<p><span style="font-size: 16px;"><a href="/default/index/url?u=aHR0cHM6Ly93d3cuY25ibG9ncy5jb20vY2t4a2V4aW5nL3AvMTEyMTk2MTIuaHRtbA==" target="_blank">https://www.cnblogs.com/ckxkexing/p/11219612.html</a></span></p>
<p> </p>
<p><a href="https://blog.csdn.net/qq_41785863/article/details/98469396" rel="nofollow" target="_blank">https://blog.csdn.net/qq_41785863/article/details/98469396</a></p>
<p><a href="/default/index/url?u=aHR0cHM6Ly93d3cuY25ibG9ncy5jb20vYnBkd24tY25ibG9ncy9wLzExMjUyMTg1Lmh0bWw=" target="_blank">https://www.cnblogs.com/bpdwn-cnblogs/p/11252185.html</a></p>
<p> </p>
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