fork download
  1. #include<bits/stdc++.h>
  2. using namespace std;
  3. #define endl '\n'
  4. #define int long long int
  5. const int MOD = 1000000007;
  6. const int MOD2 = 998244353;
  7. const int INF = LLONG_MAX / 2;
  8. const int MAXN = 100000;
  9. int primes[1000000];
  10.  
  11. /*void seive() {
  12.   fill(primes, primes + 1000000, 1);
  13.   primes[0] = primes[1] = 0;
  14.   for (int i = 2; i * i < 1000000; i++) {
  15.   if (primes[i]) {
  16.   for (int j = i * i; j < 1000000; j += i) {
  17.   primes[j] = 0;
  18.   }
  19.   }
  20.   }
  21. }
  22.  
  23. bool isPrime(int n) {
  24.   if (n <= 1) return false;
  25.   for (int i = 2; i * i <= n; i++) {
  26.   if (n % i == 0) return false;
  27.   }
  28.   return true;
  29. }
  30.  
  31. int gcd(int a, int b) {
  32.   if (a == 0) return b;
  33.   return gcd(b % a, a);
  34. }*/
  35.  
  36. int power(int a, int b, int mod) {
  37. int res = 1;
  38. a %= mod;
  39. while (b > 0) {
  40. if (b & 1) res = res * a % mod;
  41. a = a * a % mod;
  42. b >>= 1;
  43. }
  44. return res;
  45. }
  46.  
  47. // nCr % MOD for n < MOD
  48. int nCrModP(int n, int r) {
  49. if (r > n) return 0;
  50. if (r == 0 || r == n) return 1;
  51.  
  52. int numerator = 1, denominator = 1;
  53. for (int i = 0; i < r; i++) {
  54. numerator = (numerator * (n - i)) % MOD;
  55. denominator = (denominator * (i + 1)) % MOD;
  56. }
  57. return (numerator * power(denominator, MOD - 2, MOD)) % MOD;
  58. }
  59.  
  60. // Lucas's Theorem
  61. int lucas(int n, int r) {
  62. if (r == 0) return 1;
  63. return (lucas(n / MOD, r / MOD) * nCrModP(n % MOD, r % MOD)) % MOD;
  64. }
  65. void solve() {
  66. int n,k;
  67. cin>>n>>k;
  68. int A[n];
  69. int cnt = 0;
  70. for(int i = 0 ; i<n ; i++){
  71. cin>>A[i];
  72. }
  73. for(int i = 0 ; i<n ; i++){
  74. if(A[i]==1){
  75. cnt++;
  76. }
  77. }
  78. int maxi = INT_MAX,cnt1=0;
  79. for(int i = 0 ; i<k ; i++){
  80. if(A[i]==1){
  81. cnt1++;
  82. }
  83. }
  84. maxi = min(maxi,cnt1);
  85. int i = 0,j=k;
  86. while(j<n){
  87. if(A[i]==1){
  88. cnt1--;
  89. }
  90. if(A[j]==1){
  91. cnt1++;
  92. }
  93. maxi = min(maxi,cnt1);
  94. i++;
  95. j++;
  96. }
  97. int d = cnt-maxi;
  98. int e = (maxi*(maxi+1))/2;
  99. cout<<d+e<<endl;
  100. }
  101. signed main() {
  102. ios::sync_with_stdio(false); cin.tie(NULL);
  103. int t;
  104. cin >> t;
  105. while (t--) {
  106. solve();
  107. }
  108. return 0;
  109. }
  110.  
Success #stdin #stdout 0.01s 5320KB
stdin
3
5 4
1 1 1 0 1
5 3
1 0 1 0 1
4 3
1 1 0 1
stdout
7
3
4