Lightoj 1048-conquering Keokradong "two points"

Title Link: http://www.lightoj.com/volume_showproblem.php?problem=1048

Test instructions: There are N numbers, which are divided into M parts in order, M<=min (n,300). So that the maximum value of each share is minimized. There are a lot of cases in which the first day, the next day ... And the biggest case.

Idea: Two of the largest and, get a few mid. Determine if the condition is met.

Code: (No AC really can not find out, first put)

#include <stdio.h>#include <ctime>#include <math.h>#include <limits.h>#include <complex>#include <string>#include <functional>#include <iterator>#include <algorithm>#include <vector>#include <stack>#include <queue>#include <set>#include <map>#include <list>#include <bitset>#include <sstream>#include <iomanip>#include <fstream>#include <iostream>#include <ctime>#include <cmath>#include <cstring>#include <cstdio>#include <time.h>#include <ctype.h>#include <string.h>#include <assert.h>using namespace STD;inta[1010];intTintN, K;inttmpintsum[1010];intAnsintg[1010];intIS_OK (intx) {intResintnum =0;intpos =0; for(inti =1; I <= N; i++)if(Sum[i]-sum[i-1] > x)return 0; for(inti =1; I <= N; i++) {if(Sum[i]-sum[pos] = = X | | (i = = n && sum[i]-sum[pos] <= x) | | (I! = n && sum[i+1]-Sum[pos] > x)) {if(I! = n) num++;        pos = i; }    }if(Num <= k)return 1;Else return 0;}voidGet_ans () {intResintnow =1;intremain = N-k-1; while(now <= N) {intpos = Now-1;intTMP = Sum[now]-sum[now-1]; now++;if(Remain! =0)        { for(; now <= N; now++) {if(Sum[now]-sum[pos] <= ans) tmp = Sum[now]-Sum[pos], remain--;Else{printf("%d\n", TMP); Break; }            }if(Remain = =0)            {printf("%d\n", TMP); now++; Break; }        }Else{printf("%d\n", TMP); }    }}intMain () {scanf("%d", &t);intCases =1; while(t--) {scanf("%d%d", &n, &k); n++;inttot =0;memset(Sum,0,sizeof(sum)); for(inti =1; I <= N; i++) {scanf("%d", &a[i]); Sum[i] = sum[i-1] + a[i];        Tot + = A[i]; }intleft =1;intright = tot;intMid while(left <= right) {mid = (left + right)/2;if(Is_ok (mid))                {ans = mid; right = mid-1; }ElseLeft = mid +1; }printf("Case%d:", cases++);printf("%d\n", ans);    Get_ans (); }return 0; }

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Lightoj 1048-conquering Keokradong "two points"