HDU 3273 二分

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不一样的二分；

思路：

*二分的上限下限是：MAX（最大的数字），SUM （总和）；

*其中一定有一个数字满足，将N分成M份；

*求最小的；只要当它达到M份时，依然减小上届值；

算法不难，思路想到了就水到渠成了；

/*   *problem ID: POJ 3273 * Author ID: fuqiang   *      TIME: 2013-07-17   * Algorithm: 二分  *    Status: (Accept)  */  #include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <iomanip>#define maxn 100010int a[maxn],N,M;using namespace std;inline bool check(int mid){int num = 1, sum = 0;for(int i = 0; i <= N; i++){if(sum + a[i] <= mid)sum += a[i];else{sum = a[i];num++;}}return num>M;}int main(int argc, char *argv[]){while(scanf("%d%d",&N,&M)!=EOF){int MAX = 0, sum = 0;for(int i = 1; i <= N; i++){scanf("%d",&a[i]);sum += a[i];MAX = max(MAX,a[i]);}int left = MAX, right = sum, mid;while(left < right){mid = (left + right) / 2;if(check(mid))left = mid + 1;elseright = mid - 1;}printf("%d\n",left); }return 0;}