就拿样例来说，我的二分法明显告诉我只要和数是5以上，那么都可以成立，但是5不是任何组合的和，我应该如何加个条件让二分法分出来的答案正好为最小的和```c
#include<stdio.h>
#include<string.h>
#include<stdbool.h>
bool check(int mid,int arr,int n,int m)
{
int sum = 0;
for (int i = 0; i < n;i++)
{
sum += arr[i];
if(sum >= mid)
{
/ if(sum == mid)
{
flag = 1;
} */
m--;
sum = 0;
}
}
if(m>=0)
{
return true;
}
return false;
}
int main()
{
int N = 0, M = 0;
int arr[10002];
memset(arr, 0, sizeof(arr));
scanf("%d%d",&N,&M);
for (int i = 0; i < N;i++)
{
scanf("%d", &arr[i]);
}
//二分答案的查询
int left = 0;
int right = 100;
while(left < right)
{
int mid = left + (right - left) / 2;
if(check(mid,arr,N,M - 1))
{
right = mid;
}
else
{
left = mid + 1;
}
}
printf("%d",left+1);
}
