So, they ask you to write a program that checks whether there is a fair partition of the marbles. Input Each line in the input describes one collection of marbles to be divided. The lines consist of six non-neg
Each line in the input file describes one collection of marbles to be divided. The lines contain six non-negative integers n1 , . . . , n6 , where ni is the number of marbles of value i. So, the example from above would be described by the input-line "1 0 1 2 0 0". The max...
输入 Each line in the input file describes one collection of marbles to be divided. The lines contain six non-negative integers n1 , . . . , n6 , where ni is the number of marbles of value i. So, the example from above would be described by the input-line "1 0 1 2 0 0". Th...
1. In the companion paper Bogomolnaia et al. (2017) we consider the more general problem of dividing a “mixed manna” containing both goods and bads, as when we dissolve a partnership with both valuable assets and liabilities. Our first observation is that the E rule is no longer well d...
0149-Max-Points-on-a-Line 0150-Evaluate-Reverse-Polish-Notation 0151-Reverse-Words-in-a-String 0152-Maximum-Product-Subarray 0153-Find-Minimum-in-Rotated-Sorted-Array 0155-Min-Stack 0159-Longest-Substring-with-At-Most-Two-Distinct-Characters 0160-Intersection-of-Two-Link...
#include<iostream>#include<algorithm>usingnamespacestd;intn[7];intresult[120001];boolisSuccess;intmaxSize;intmaxNum;voidstart(intlast,intstartPos){maxNum+=n[startPos]*startPos;for(inti=min(maxNum,last);i>=1;i--){for(intj=0;j<=n[startPos]&&j*startPos<=last;j++){if(j*startPos+result...
#include<algorithm> #include<queue> #include<vector> #include<cmath> #include<set> #include<cstdlib> #include<cstring> #include<stack> #include<string> using namespace std; //freopen("C://i.txt","r",stdin); #define N 1000111
#include<iostream>#include<stdio.h>#include<stdlib.h>#include<iomanip>#include<cmath>#include<float.h>#include<string.h>#include<algorithm>#definesf scanf#definepf printf#definescf(x) scanf("%d",&x)#definescff(x,y) scanf("%d%d",&x,&y)#defineprf(x) printf("%d\n",x)#definemm(x...
Output a blank line after each test case. Sample Input 1 0 1 2 0 0 1 0 0 0 1 1 0 0 0 0 0 0 Sample Output Collection #1: Can't be divided. Collection #2: Can be divided. 看看能否凑出V/2这个值就可以了。母函数比较好理解 ...
#include <algorithm> using namespace std; int a[100],cases=0,dp[666666],f[666666],tot,all,w; bool flag; int main() { while(1) { cases++; tot=0;flag=false;all=0; memset(dp,0,sizeof(dp)); memset(f,0,sizeof(f));