CodeForces 1197D Yet Another Subarray Problem

Time limit 2000 ms

Memory limit 262144 kB

Source Educational Codeforces Round 69 (Rated for Div. 2)

Tags dp greedy math *1900

Editorial Announcement (en) Tutorial #1 (en) Tutorial #2 (en) Tutorial #3 (ru)

##官方题解

<div>At first let's solve this problem when <mi>m</mi><mo>=</mo><mn>1</mn></math>" role="presentation">m=1<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi><mo>=</mo><mn>1</mn></math><script type="math/tex" id="MathJax-Element-123">m = 1</script> and <mi>k</mi><mo>=</mo><mn>0</mn></math>" role="presentation">k=0<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>k</mi><mo>=</mo><mn>0</mn></math><script type="math/tex" id="MathJax-Element-124">k = 0</script> (it is the problem of finding subarray with maximum sum). For each position from <mn>1</mn></math>" role="presentation">1<math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn></math><script type="math/tex" id="MathJax-Element-125">1</script> to <mi>n</mi></math>" role="presentation">n<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>n</mi></math><script type="math/tex" id="MathJax-Element-126">n</script> we want to know the value of <mi>m</mi><mi>a</mi><mi>x</mi><msub><mi>l</mi><mi>i</mi></msub><mo>=</mo><munder><mo form="prefix">max</mo><mrow class="MJX-TeXAtom-ORD"><mn>1</mn><mo>&#x2264;</mo><mi>j</mi><mo>&#x2264;</mo><mi>i</mi><mo>+</mo><mn>1</mn></mrow></munder><mi>s</mi><mi>u</mi><mi>m</mi><mo stretchy="false">(</mo><mi>j</mi><mo>,</mo><mi>i</mi><mo stretchy="false">)</mo></math>" role="presentation">maxli=max1≤j≤i+1sum(j,i)<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi><mi>a</mi><mi>x</mi><msub><mi>l</mi><mi>i</mi></msub><mo>=</mo><munder><mo form="prefix">max</mo><mrow class="MJX-TeXAtom-ORD"><mn>1</mn><mo>≤</mo><mi>j</mi><mo>≤</mo><mi>i</mi><mo>+</mo><mn>1</mn></mrow></munder><mi>s</mi><mi>u</mi><mi>m</mi><mo stretchy="false">(</mo><mi>j</mi><mo>,</mo><mi>i</mi><mo stretchy="false">)</mo></math><script type="math/tex" id="MathJax-Element-127">maxl_i = \max\limits_{1 \le j \le i + 1} sum(j, i)</script>, where <mi>s</mi><mi>u</mi><mi>m</mi><mo stretchy="false">(</mo><mi>l</mi><mo>,</mo><mi>r</mi><mo stretchy="false">)</mo><mo>=</mo><munderover><mo movablelimits="false">&#x2211;</mo><mrow class="MJX-TeXAtom-ORD"><mi>k</mi><mo>=</mo><mi>l</mi></mrow><mrow class="MJX-TeXAtom-ORD"><mi>k</mi><mo>&#x2264;</mo><mi>r</mi></mrow></munderover><msub><mi>a</mi><mi>k</mi></msub></math>" role="presentation">sum(l,r)=∑k=lk≤rak<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>s</mi><mi>u</mi><mi>m</mi><mo stretchy="false">(</mo><mi>l</mi><mo>,</mo><mi>r</mi><mo stretchy="false">)</mo><mo>=</mo><munderover><mo movablelimits="false">∑</mo><mrow class="MJX-TeXAtom-ORD"><mi>k</mi><mo>=</mo><mi>l</mi></mrow><mrow class="MJX-TeXAtom-ORD"><mi>k</mi><mo>≤</mo><mi>r</mi></mrow></munderover><msub><mi>a</mi><mi>k</mi></msub></math><script type="math/tex" id="MathJax-Element-128">sum(l, r) = \sum\limits_{k = l}^{k \le r} a_k</script>, and <mi>s</mi><mi>u</mi><mi>m</mi><mo stretchy="false">(</mo><mi>x</mi><mo>+</mo><mn>1</mn><mo>,</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mn>0</mn></math>" role="presentation">sum(x+1,x)=0<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>s</mi><mi>u</mi><mi>m</mi><mo stretchy="false">(</mo><mi>x</mi><mo>+</mo><mn>1</mn><mo>,</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mn>0</mn></math><script type="math/tex" id="MathJax-Element-129">sum(x+1, x) = 0</script>. We will calculate it the following way. <mi>m</mi><mi>a</mi><mi>x</mi><msub><mi>l</mi><mi>i</mi></msub></math>" role="presentation">maxli<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>m</mi><mi>a</mi><mi>x</mi><msub><mi>l</mi><mi>i</mi></msub></math><script type="math/tex" id="MathJax-Element-130">maxl_i</script> will be the maximum of two values: <ul> <li> <mn>0</mn></math>" role="presentation">0<math xmlns="http://www.w3.org/1998/Math/MathML"><mn>0</mn></math><script type="math/tex" id="MathJax-Element-131">0</script> (because we can take segments of length <mn>0</mn></math>" role="presentation">0<math xmlns="http://www.w3.org/1998/Math/MathML"><mn>0</mn></math><script type="math/tex" id="MathJax-Element-132">0</script>); </li><li> <msub><mi>a</mi><mi>i</mi></msub><mo>+</mo><mi>m</mi><mi>a</mi><mi>x</mi><msub><mi>l</mi><mrow class="MJX-TeXAtom-ORD"><mi>i</mi><mo>&#x2212;</mo><mn>1</mn></mrow></msub></math>" role="presentation">ai+maxli−1<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>a</mi><mi>i</mi></msub><mo>+</mo><mi>m</mi><mi>a</mi><mi>x</mi><msub><mi>l</mi><mrow class="MJX-TeXAtom-ORD"><mi>i</mi><mo>−</mo><mn>1</mn></mrow></msub></math><script type="math/tex" id="MathJax-Element-133">a_i + maxl_{i-1}</script>. </li></ul>The maximum sum of some subarray is equal to <munder><mo form="prefix">max</mo><mrow class="MJX-TeXAtom-ORD"><mn>1</mn><mo>&#x2264;</mo><mi>i</mi><mo>&#x2264;</mo><mi>n</mi></mrow></munder><mi>m</mi><mi>a</mi><mi>x</mi><msub><mi>l</mi><mi>i</mi></msub></math>" role="presentation">max1≤i≤nmaxli<math xmlns="http://www.w3.org/1998/Math/MathML"><munder><mo form="prefix">max</mo><mrow class="MJX-TeXAtom-ORD"><mn>1</mn><mo>≤</mo><mi>i</mi><mo>≤</mo><mi>n</mi></mrow></munder><mi>m</mi><mi>a</mi><mi>x</mi><msub><mi>l</mi><mi>i</mi></msub></math><script type="math/tex" id="MathJax-Element-134">\max\limits_{1\le i \le n} maxl_i</script>.So, now we can calculate the values of <mi>b</mi><mi>e</mi><mi>s</mi><msub><mi>t</mi><mi>i</mi></msub><mo>=</mo><munder><mo form="prefix">max</mo><mrow class="MJX-TeXAtom-ORD"><mn>0</mn><mo>&#x2264;</mo><mi>l</mi><mi>e</mi><mi>n</mi><mo>,</mo><mi>i</mi><mo>&#x2212;</mo><mi>l</mi><mi>e</mi><mi>n</mi><mo>&#x22C5;</mo><mi>m</mi><mo>&#x2265;</mo><mn>0</mn></mrow></munder><mo stretchy="false">(</mo><mi>s</mi><mi>u</mi><mi>m</mi><mo stretchy="false">(</mo><mi>i</mi><mo>&#x2212;</mo><mi>l</mi><mi>e</mi><mi>n</mi><mo>&#x22C5;</mo><mi>m</mi><mo>+</mo><mn>1</mn><mo>,</mo><mi>i</mi><mo stretchy="false">)</mo><mo>&#x2212;</mo><mi>l</mi><mi>e</mi><mi>n</mi><mo>&#x2217;</mo><mi>k</mi><mo stretchy="false">)</mo></math>" role="presentation">besti=max0≤len,i−len⋅m≥0(sum(i−len⋅m+1,i)−len∗k)<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>b</mi><mi>e</mi><mi>s</mi><msub><mi>t</mi><mi>i</mi></msub><mo>=</mo><munder><mo form="prefix">max</mo><mrow class="MJX-TeXAtom-ORD"><mn>0</mn><mo>≤</mo><mi>l</mi><mi>e</mi><mi>n</mi><mo>,</mo><mi>i</mi><mo>−</mo><mi>l</mi><mi>e</mi><mi>n</mi><mo>⋅</mo><mi>m</mi><mo>≥</mo><mn>0</mn></mrow></munder><mo stretchy="false">(</mo><mi>s</mi><mi>u</mi><mi>m</mi><mo stretchy="false">(</mo><mi>i</mi><mo>−</mo><mi>l</mi><mi>e</mi><mi>n</mi><mo>⋅</mo><mi>m</mi><mo>+</mo><mn>1</mn><mo>,</mo><mi>i</mi><mo stretchy="false">)</mo><mo>−</mo><mi>l</mi><mi>e</mi><mi>n</mi><mo>∗</mo><mi>k</mi><mo stretchy="false">)</mo></math><script type="math/tex" id="MathJax-Element-135">best_i = \max\limits_{0 \le len, i - len \cdot m \ge 0} (sum(i-len \cdot m + 1, i) - len * k)</script> the same way. <mi>b</mi><mi>e</mi><mi>s</mi><msub><mi>t</mi><mi>i</mi></msub></math>" role="presentation">besti<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>b</mi><mi>e</mi><mi>s</mi><msub><mi>t</mi><mi>i</mi></msub></math><script type="math/tex" id="MathJax-Element-136">best_i</script> is the maximum of two values: <ul> <li> 0; </li><li> <mi>s</mi><mi>u</mi><mi>m</mi><mo stretchy="false">(</mo><mi>i</mi><mo>&#x2212;</mo><mi>m</mi><mo>+</mo><mn>1</mn><mo>,</mo><mi>i</mi><mo stretchy="false">)</mo><mo>&#x2212;</mo><mi>k</mi><mo>+</mo><mi>b</mi><mi>e</mi><mi>s</mi><msub><mi>t</mi><mrow class="MJX-TeXAtom-ORD"><mi>i</mi><mo>&#x2212;</mo><mi>m</mi></mrow></msub></math>" role="presentation">sum(i−m+1,i)−k+besti−m<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>s</mi><mi>u</mi><mi>m</mi><mo stretchy="false">(</mo><mi>i</mi><mo>−</mo><mi>m</mi><mo>+</mo><mn>1</mn><mo>,</mo><mi>i</mi><mo stretchy="false">)</mo><mo>−</mo><mi>k</mi><mo>+</mo><mi>b</mi><mi>e</mi><mi>s</mi><msub><mi>t</mi><mrow class="MJX-TeXAtom-ORD"><mi>i</mi><mo>−</mo><mi>m</mi></mrow></msub></math><script type="math/tex" id="MathJax-Element-137">sum(i - m + 1, i) - k + best_{i-m}</script>. </li></ul>After calculating all values <mi>b</mi><mi>e</mi><mi>s</mi><msub><mi>t</mi><mi>i</mi></msub></math>" role="presentation">besti<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>b</mi><mi>e</mi><mi>s</mi><msub><mi>t</mi><mi>i</mi></msub></math><script type="math/tex" id="MathJax-Element-138">best_i</script> we can easily solve this problem. At first, let's iterate over the elements <mi>b</mi><mi>e</mi><mi>s</mi><msub><mi>t</mi><mi>i</mi></msub></math>" role="presentation">besti<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>b</mi><mi>e</mi><mi>s</mi><msub><mi>t</mi><mi>i</mi></msub></math><script type="math/tex" id="MathJax-Element-139">best_i</script>. When we fix some element <mi>b</mi><mi>e</mi><mi>s</mi><msub><mi>t</mi><mi>i</mi></msub></math>" role="presentation">besti<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>b</mi><mi>e</mi><mi>s</mi><msub><mi>t</mi><mi>i</mi></msub></math><script type="math/tex" id="MathJax-Element-140">best_i</script>, lets iterate over the value <mi>l</mi><mi>e</mi><mi>n</mi><mo>=</mo><mn>1</mn><mo>,</mo><mn>2</mn><mo>,</mo><mo>&#x2026;</mo><mo>,</mo><mi>m</mi></math>" role="presentation">len=1,2,…,m<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>l</mi><mi>e</mi><mi>n</mi><mo>=</mo><mn>1</mn><mo>,</mo><mn>2</mn><mo>,</mo><mo>…</mo><mo>,</mo><mi>m</mi></math><script type="math/tex" id="MathJax-Element-141">len = 1, 2, \dots, m</script> and update the answer with value <mi>b</mi><mi>e</mi><mi>s</mi><msub><mi>t</mi><mi>i</mi></msub><mo>+</mo><mi>s</mi><mi>u</mi><mi>m</mi><mo stretchy="false">(</mo><mi>i</mi><mo>&#x2212;</mo><mi>l</mi><mi>e</mi><mi>n</mi><mo>,</mo><mi>i</mi><mo>&#x2212;</mo><mn>1</mn><mo stretchy="false">)</mo><mo>&#x2212;</mo><mi>k</mi></math>" role="presentation">besti+sum(i−len,i−1)−k<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>b</mi><mi>e</mi><mi>s</mi><msub><mi>t</mi><mi>i</mi></msub><mo>+</mo><mi>s</mi><mi>u</mi><mi>m</mi><mo stretchy="false">(</mo><mi>i</mi><mo>−</mo><mi>l</mi><mi>e</mi><mi>n</mi><mo>,</mo><mi>i</mi><mo>−</mo><mn>1</mn><mo stretchy="false">)</mo><mo>−</mo><mi>k</mi></math><script type="math/tex" id="MathJax-Element-142">best_i + sum(i - len, i - 1) - k</script>.</div>

##源代码

#include<stdio.h>
#include<algorithm>

int n,m,k;
long long a[300010];
long long dp[300010],ans;
int main()
{
	//freopen("test.in","r",stdin);
	scanf("%d%d%d",&n,&m,&k);
	for(int i=1;i<=n;i++) scanf("%lld",a+i),a[i]+=a[i-1];
	for(int i=1;i<=n;i++)
	{
		for(int j=i;j+m>=i;j--)
			dp[i]=std::max(dp[i],a[i]-a[j]);
		dp[i]-=k;
		dp[i]=std::max(0LL,dp[i]);
		if(i>m) dp[i]=std::max(dp[i],dp[i-m]+a[i]-a[i-m]-k);
		ans=std::max(dp[i],ans);
	}
	printf("%lld\n",ans);
	return 0;
}

来源：oschina

链接：https://my.oschina.net/u/4388412/blog/3452567

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