HDU5086Revenge of Segment Tree(数论)

pid=5086" target="_blank" style="">题目链接

题目大意:给出长度为n的数组。然后要求累计里面的每一个子串的和。

解题思路:枚举起点和终点,推断每一个数属于多少条线段。那么这个数就要被加这么多次。能够得出每一个位置被加次数的公式: i
(n - i + 1);那么结果就是累计(arr[i] i) mod * (n - i + 1) % mod,注意两个数相乘可能会超出long long型的数。

代码:

#include <cstdio>
#include <cstring> typedef long long ll; const int maxn = 1e6;
const ll mod = 1e9 + 7; ll arr[maxn]; int main () { int T, n;
scanf ("%d", &T);
while (T--) { scanf ("%d", &n);
for (int i = 1; i <= n; i++)
scanf ("%I64d", &arr[i]); ll ans = 0;
for (int i = 1; i <= n; i++)
ans = (ans + (arr[i] * i % mod) * (n - i + 1) % mod) % mod;
printf ("%I64d\n", ans);
}
return 0;
}

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