HDU 3273 二分

不一樣的二分；

思路：

    *二分的上限下限是：MAX（最大的數字），SUM （總和）；

    *其中一定有一個數字滿足，將N分成M份；

    *求最小的；只要當它達到M份時，依然減小上屆值；

演算法不難，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;}