提出

src/apple.c

@@ -1,17 +1,26 @@
 #include <stdio.h>
 
-int n;
-int k;
+int n = 0;
+int k = 0;
 int A[100000];
 
+int p(int i1,int i2,int X[]){
+  return X[i2] > i1;
+}
 
 int main(){
-  int i, lb, ub;
+  int i, lb, ub = 0;
   scanf("%d%d", &n, &k);
   for(i = 0; i < n; i++){
     scanf("%d", &A[i]);
   }
-
-
+  lb = -1;
+  ub = n;
+  while(ub - lb > 1){
+    int mid = (lb + ub) / 2;
+    if(p(k,mid,A)) ub = mid;
+    else lb = mid;
+  }
+  printf("%d\n", ub);
   return 0;
 }

src/array.c

@@ -4,15 +4,29 @@ int n;
 int k;
 int A[100000];
 
+int p(int i1, int i2, int i3,int X[]){
+  int sum = i1;
+  for(int m = 0; m < i1; m++){
+    sum += (A[m] - 1) / i3;
+  }
+  return  sum > i2;
+}
 
 int main(){
   int i, lb, ub;
   scanf("%d%d", &n, &k);
   for(i = 0; i < n; i++){
     scanf("%d", &A[i]);
   }
-
-
+  lb = 0;
+  ub = 10000;
+  while(ub - lb >1){
+    int mid = (lb + ub ) / 2;
+
+    if(p(n,k,mid,A) ) lb = mid;
+    else ub = mid;
+  }
+  printf("%d\n", ub);
 
   return 0;
 }

src/spear.c

@@ -4,14 +4,28 @@ int n;
 int k;
 int A[100000];
 
+int p(int i1,int i2, int i3, int X[]){
+  int sum = 0;
+  for(int m = 0;m < i1; m++){
+    sum += X[m] / i3;
+  }
+  return sum < i2;
+}
 
 int main(){
   int i, lb, ub;
   scanf("%d%d", &n, &k);
   for(i = 0; i < n; i++){
     scanf("%d", &A[i]);
   }
-
+   lb = 0;
+  ub = 1000000000;
+  while(ub - lb >1){
+    int mid = (lb + ub ) / 2;
+    if(p(n,k,mid,A)) ub = mid;
+    else lb = mid;
+  }
+  printf("%d\n", lb);
 
   return 0;
 }

src/works.c

@@ -4,14 +4,42 @@ int n;
 int k;
 int A[100000];
 
+int p(int l1,int l2,int l3, int X[]){
+    int i1 = 0;
+    int i2 = 0;
+    int sum = 0;
+    while(i2<l1){
+        sum += X[i2];
+        if(sum + X[i2 + 1] < l3){
+            i2++;
+        }    
+        else if(sum < l3 && sum + A[i2 + 1] > l3){
+            sum = 0;
+            i1++;
+            i2++;
+        }
+        else return 0;
+    }
+    return i1 <= l2; 
+}
 
 int main(){
   int i, lb, ub;
   scanf("%d%d", &n, &k);
   for(i = 0; i < n; i++){
     scanf("%d", &A[i]);
   }
-
-
+  int u = 0;
+  for(int i = 0;i<n;i++){
+      u += A[i];
+  }
+  lb = 1;
+  ub = 1000000000;
+  while(ub - lb >1){
+    int mid = ( lb + ub ) / 2;
+    if(p(n,k,mid,A)) ub = mid;
+    else lb = mid;
+  }
+  printf("%d\n", lb);
   return 0;
 }