uva 714 Copying Books(二分法求最大值最小化）

题目连接：714 - Copying Books

题目大意：将一个个数为n的序列分割成m份，要求这m份中的每份中值（该份中的元素和）最大值最小， 输出切割方式，有多种情况输出使得越前面越小的情况。

解题思路：二分法求解f(x), f(x) < 0 说明不能满足， f(x) >= 0说明可以满足，f(x) 就是当前最大值为x的情况最少需要划分多少份-要求份数（如果f(x ) >= 0 说明符合要求而且还过于满足，即x还可以更小）。

注意用long long .

 

#include <stdio.h>

#include <string.h>

int max(const int &a, const int &b) { return a > b ? a : b;}



const int N = 1005;

int n, T;

long long num[N], sum[N], rec[N];



bool judge(int Max) {

    int cnt = 0, first = 0, end = 1;

    while (end <= n) {

	if (sum[end] - sum[first] > Max) {

	    cnt++;

	    first = end - 1;

	}

	end++;

    }

    return cnt <= T - 1;

}



int main () {

    long long  cas, lift, right, cur;

    scanf("%lld", &cas);

    while (cas--) {

	// Init;

	memset(num, 0, sizeof(num));

	memset(sum, 0, sizeof(sum));

	memset(rec, 0, sizeof(rec));

	lift = right = 0;



	// Read.

	scanf("%d%d", &n, &T);

	for (int i = 1; i <= n; i++) {

	    scanf("%lld", &num[i]);

	    sum[i] = sum[i - 1] + num[i];

	    lift = max(lift, num[i]);

	}

	right = sum[n];



	// Count;

	while (lift != right) {

	    cur = (right + lift) / 2;

	    if (judge(cur)) {

		right = cur;

	    }

	    else

		lift = cur + 1;

	}



	// Draw;

	long long now = 0, cnt = 0;

	for (int i = n; i > 0; i--) {

	    if (now + num[i] > lift || i < T - cnt) {

		rec[i] = 1;

		cnt++;

		now = num[i];

	    }

	    else

		now += num[i];

	}



	// Printf;

	for (int i = 1; i < n; i++) {

	    printf("%lld ", num[i]);

	    if (rec[i])	    printf("/ ");

	}

	printf("%lld\n", num[n]);

    }

    return 0;

}