提出 二分探索

src/apple.c

@@ -4,14 +4,37 @@ int n;
 int k;
 int A[100000];
 
+//各成分を推定値で割ったときの切り上げの値を足す
+int p(int m){
+  int sum = 0;
+  for(int j = 0; j < n; j++){
+    sum += (A[j] + m - 1) / m;
+  }
+  return (long long int) sum <= k;
+}
+
+//ubの値に代入するために配列の最大値を求める
+int m(int B[]){
+  int max = 0;
+  for(int j = 0; j < n; j++){
+    if(max <= B[j]) max = B[j];
+  }
+  return max;
+}
 
 int main(){
   int i, lb, ub;
   scanf("%d%d", &n, &k);
   for(i = 0; i < n; i++){
     scanf("%d", &A[i]);
   }
-
-
+  lb = 0;
+  ub = m(A);
+  while(ub - lb > 1){
+    int mid = (lb + ub) / 2;
+    if(p(mid)) ub = mid;
+    else lb = mid;
+  }
+  printf("%d\n", ub);
   return 0;
 }

src/array.c

@@ -11,8 +11,13 @@ int main(){
   for(i = 0; i < n; i++){
     scanf("%d", &A[i]);
   }
-
-
-
+  lb = -1;
+  ub = n;
+  while(ub - lb > 1){
+    int mid = (lb + ub)/2;
+    if(A[mid] >= k) ub = mid;
+    else lb = mid;
+  }
+  printf("%d\n", ub);
   return 0;
 }

src/spear.c

@@ -4,14 +4,38 @@ int n;
 int k;
 int A[100000];
 
+//各成分を推定値で割ったときの切り捨ての値を足す
+int p(int m){
+  int sum = 0;
+  for(int j = 0; j < n; j++){
+    if(m != 0) sum += A[j] / m;
+    else sum = k + 1;
+  }
+  return (long long int) sum < k;
+}
+
+//ubの値に代入するために配列の成分の和を求める
+int m(int B[]){
+  int max = 0;
+  for(int j = 0; j < n; j++){
+    if(max <= B[j]) max = B[j];
+  }
+  return max;
+}
 
 int main(){
   int i, lb, ub;
   scanf("%d%d", &n, &k);
   for(i = 0; i < n; i++){
     scanf("%d", &A[i]);
   }
-
-
+  lb = 0;
+  ub = m(A);
+  while(ub - lb > 1){
+    int mid = (ub + lb) / 2;
+    if(p(mid)) ub = mid;
+    else lb = mid;
+  }
+  printf("%d\n", lb);
   return 0;
 }

src/works.c

@@ -1,17 +1,54 @@
 #include <stdio.h>
+#include <stdbool.h>
+#include <string.h>
 
-int n;
-int k;
-int A[100000];
+unsigned int n;
+unsigned int k;
+unsigned int A[100000];
 
 
+int s(){
+  int sum = 0;
+  for(int j = 0; j < n; j++){
+    sum += A[j];
+  }
+  return sum;
+}
+
+int m(unsigned int B[]){
+  int max = 0;
+  for(int j = 0; j < n; j++){
+    if(max <= B[j]) max = B[j];
+  }
+  return max;
+}
+
+//人の入れ替わりがk-1回起こるようにする
+int p(int m){
+  int people = 1;
+  for(int a = 0; a < n-1; a++){
+    if(A[a] + A[a+1] <= m){
+      A[a+1] = A[a] + A[a+1];
+    }
+    else people++;
+  }
+  return (long long int) people <= k;
+}
+
 int main(){
   int i, lb, ub;
   scanf("%d%d", &n, &k);
   for(i = 0; i < n; i++){
     scanf("%d", &A[i]);
   }
-
-
+  lb = m(A) - 1;
+  ub = s();
+  while(ub - lb > 1){
+    int mid = (lb + ub) / 2;
+    if(p(mid)) ub = mid;
+    else lb = mid;
+  }
+
+  printf("%d\n", ub);
   return 0;
 }