BestCoder Round #85 sum
大晚上的更一道下午的水题吧。(虽然WA了好多次= =，但真实情况是我比较水)
描述
Given a sequence, you're asked whether there exists a consecutive subsequence whose sum is divisible by m. output YES, otherwise output NO.
输入
The first line of the input has an integer T (1≤T≤10), which represents the number of test cases. For each test case, there are two lines: 1.The first line contains two positive integers n, m (1≤n≤100000, 1≤m≤5000). 2.The second line contains n positive integers x (1≤x≤100) according to the sequence.
输出
Output T lines, each line print a YES or NO.
样例输入
2 3 3 1 2 3 5 7 6 6 6 6 6
样例输出
YES NO
代码如下：
1 #include <cstdio> 2 #define N 100100 3 4 int main() 5 { 6 int t,n,m,a[N]; 7 int flag=0; 8 scanf("%d",&t); 9 while(t--){ 10 scanf("%d %d",&n,&m); 11 int sum=0; 12 int k=1; 13 for(int j=1;j<=n;j++){ 14 scanf("%d",&a[j]); 15 } 16 for(int i=1;i<=n;i++){ 17 while(k<=n&&sum<m){ 18 sum+=a[k++]; 19 } 20 if(sum==m){ 21 flag=1; 22 break; 23 } 24 sum-=a[i]; 25 26 } 27 if(flag==1){ 28 printf("YES\n"); 29 }else 30 printf("NO\n"); 31 flag=0; 32 } 33 return 0; 34 }
题意：
题意：能不能找一个连续子区间的和是m的倍数。
思路总结：
简单尺取法。既然说找的是m的倍数，那就从头开始加起来。加到数小于m且加起来得到的和是最大的小于m的数。开始判断。如果失败就从头开始减一个，在接着从后加一个。一点点枚举。像尺子一样~~所以叫尺取法~~HOHO~
