You are given a sequence of integers, A1,A2,...,An. And you are allowed a manipulation on the sequence to transform the origin sequence into another sequenceB1,B2,...,Bn(Maybe the two sequences are same ). The manipulation is specified as the following three steps:
1.Select an integer Ai and choose an arbitrary positive integer delta as you like.
2.Select some integers Aj satisfying j < i, let's suppose the selected integers are Ak1,Ak2,...,Akt , then subtract an arbitrary positive integer Di from Aki (1 ≤ i ≤ t) as long as sum(Di) = delta.
3.Subtract delta from Ai.
The manipulation can be performed any times. Can you find a way to transform A1,A2,...,An to B1,B2,...,Bn ?
Input
The input consist of multiple cases. Cases are about 100 or so. For each case, the first line contains an integer N(1 ≤ N ≤ 10000) indicating the number of the sequence. Then followed by N lines, ith line contains two integers Ai and Bi (0 ≤ Bi ≤ Ai ≤ 4294967296).
Output
Output a single line per case. Print "YES" if there is a certain way to transform Sequence A into Sequence B. Print "NO" if not.
Sample Input
33 24 25 232 07 13 1
Sample Output
YESNO
分析：将原题转化为直接在差数组上操作。利用贪心的思路，若目前节余的值t+s[i-1]仍然<c[i]，则NO。一直没有NO就是YES。
#include#include typedef long long u64;u64 a[10010], b[10010], c[10010], s[10010], t;int n;int main() { int i; while (scanf("%d", &n) != EOF) { s[0] = 0; t = 0; for (i = 1; i <= n; ++i) { scanf("%lld%lld", &a[i], &b[i]); c[i] = a[i] - b[i]; s[i] = s[i - 1] + c[i]; } if (s[n]&1) { printf("NO\n"); continue; } for (i = n; i >= 1; --i) { if (s[i - 1] + t < c[i]) { printf("NO\n"); goto L; } if (s[i - 1] - c[i] <= t) { break; } t += c[i]; } printf("YES\n");L: ; } return 0;}