题目:http://poj.org/problem?id=3273
题意:把n个数分成m份,使每份的和尽量小,输出最大的那一个的和。
思路:二分枚举最大的和,时间复杂度为O(nlog(sum-max));
一道很好的题。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <algorithm> 6 using namespace std; 7 const int maxn = 100000+10; 8 int a[maxn]; 9 10 int main() 11 { 12 int n, m, i, Max, sum; 13 while(~scanf("%d%d", &n, &m)) 14 { 15 Max = -1; 16 sum = 0; 17 for(i = 0; i < n; i++) 18 { 19 scanf("%d", &a[i]); 20 sum += a[i]; 21 if(Max < a[i]) 22 Max = a[i]; 23 } 24 int high = sum, low = Max, mid; 25 while(high>low) 26 { 27 mid = (high+low)/2; 28 int cou = 1, w = 0; 29 for(i = 0; i < n; i++) 30 { 31 w += a[i]; 32 if(w>mid) 33 { 34 cou++; //cou记录最大值为mid的情况下,可以分成几份 35 w = a[i]; 36 } 37 } 38 if(cou>m) 39 low = mid+1; 40 else 41 high = mid-1; 42 } 43 printf("%d ", high); 44 } 45 return 0; 46 }
以后还是用这种形式的二分吧
while(left<right)
{
if()
left=mid+1;
else right=mid;
}
代码:
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <algorithm> 6 using namespace std; 7 const int maxn = 100000+10; 8 int a[maxn]; 9 10 int main() 11 { 12 int n, m, i, Max, sum; 13 while(~scanf("%d%d", &n, &m)) 14 { 15 Max = -1; 16 sum = 0; 17 for(i = 0; i < n; i++) 18 { 19 scanf("%d", &a[i]); 20 sum += a[i]; 21 if(Max < a[i]) 22 Max = a[i]; 23 } 24 int high = sum, low = Max, mid; 25 while(high > low) 26 { 27 mid = (high+low)/2; 28 int cou = 1, w = 0; 29 for(i = 0; i < n; i++) 30 { 31 w += a[i]; 32 if(w>mid) 33 { 34 cou++; 35 w = a[i]; 36 } 37 } 38 if(cou <= m) 39 high = mid; 40 else 41 low = mid+1; 42 } 43 printf("%d ", low); 44 } 45 return 0; 46 }