lightoj1048 Conquering Keokradong

这是一个很不错的二分，，，可是wa了半天没做对，主要是二分最大单个区间的值，，，mid，

当a[i] > mid的时候，low＋＋，当cnt > k时，low＋＋；cnt <= k时，high ＝ mid；

然后就是打印比较容易挂掉；注意下就好了。

// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <limits.h>
// #define DEBUG
#ifdef DEBUG
#define debug(...) printf( __VA_ARGS__ )
#else
#define debug(...)
#endif
#define MEM(x,y) memset(x, y,sizeof x)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> ii;
const int inf = 1 << 30;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[1010];
int n, k;
bool check(int x){
// cout << "x = " << x << endl;
int sum = 0,cnt = 0;
for (int i = 0;i <= n;++i){
sum += a[i];
if (sum > x){//sum = 0 || sum > 0;
cnt++;
sum -= a[i];
// printf("sum = %d\t",sum);
if (sum <= 0) return false;//a[i] > x;->low = mid + 1;
sum = a[i];
}
}
// cout << "cnt = " << cnt << endl;
return cnt <= k;//cnt > k,->low = mid + 1;
}
int solve(int low,int high){
int mid;
while(low < high){
mid = (low + high) / 2;
if (check(mid)) high = mid;//保存结果，，，cnt <= k;
else low = mid + 1;
}
return high;
}
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int t, icase = 0;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&k);
int low = 0,high = 0;
for (int i = 0;i <= n;++i){
scanf("%d",&a[i]);
low = max(low, a[i]);
high += a[i];
}
int ans = solve(low,high);
printf("Case %d: %d\n",++icase,ans);
int tot,i,j;
for (i = 0,j = 0,tot = 0;i <= n;++i){
tot += a[i];
if (tot > ans || k - j > n - i){
tot -= a[i];
printf("%d\n",tot);
tot = a[i];
j++;
}
}
printf("%d\n",tot);
}
return 0;
}