喫糖果
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 32022 Accepted Submission(s): 9106
其它種類的糖的數量和爲sum。只要滿足m-sum<=1即可,輸出Yes,否則輸出No.
思想與
bzoj 2456 mode相似
My solution:
/*2016.3.13*/
#include<stdio.h>
#include<algorithm>
using namespace std;
long long x[1000080];
int main()
{
int i,j,n,m,t;
long long sum;
scanf("%d",&t);
while(t--)
{
sum=0;
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%I64d",&x[i]);
sort(x,x+n);
for(i=0;i<n-1;i++)
sum+=x[i];
if((x[n-1]-sum)<=1)
printf("Yes\n");
else
printf("No\n");
}
return 0;
}