玲珑学院-1014-Absolute Defeat【思维】

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1014 - Absolute Defeat

Time Limit：2s Memory Limit：64MByte

Submissions：262Solved：76

DESCRIPTION

Eric has an array of integers $a_1,a_2,...,a_n$. Every time, he can choose a contiguous subsequence of length $k$ and increase every integer in the contiguous subsequence by $1$.He wants the minimum value of the array is at least $m$. Help him find the minimum number of operations needed.

INPUT

There are multiple test cases. The first line of input contains an integer$T$, indicating the number of test cases. For each test case:The first line contains three integers$n$,$m$ and $k$$(1\le n\le {10}^5,1\le k\le n,1\le m\le {10}^4)$.The second line contains $n$ integers $a_1,a_2,...,a_n$$(1\le a_i\le {10}^4)$.

OUTPUT

For each test case, output an integer denoting the minimum number of operations needed.

SAMPLE INPUT

3

2 2 2

1 1

5 1 4

1 2 3 4 5

4 10 3

1 2 3 4

SAMPLE OUTPUT

1

0

15

题意：给你一个长度为 n 的序列，然后你每次只能处理长度为 k 子序列，每次处理使这长度为 k 的子序列每个值加 1，问最少要处理几次才能使整个序列的每个元素都大于 m

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;const int MAX=1e5+10;int n,m,k;int a[2*MAX];int main(){int t;scanf("%d",&t);while(t--){scanf("%d %d %d",&n,&m,&k);memset(a,0,sizeof(a));int x,ans=0;for(int i=1;i<=n;i++){scanf("%d",&x);a[i]+=a[i-1];if(x+a[i]<m) // 每次的 a[i]只有两个值要么是 n*k，要么是 0 {int step=m-a[i]-x;ans+=step;a[i]+=step;a[i+k]-=step; // 每次处理的长度只有 k，而上面还有个累加，所以这里先把第 i+k位减 step， }      // 这样就保证了上面判断的 a[i],只可能是 两种值 }printf("%d\n",ans);}return 0;}