本文整理汇总了C++中Answer::normalize方法的典型用法代码示例。如果您正苦于以下问题：C++ Answer::normalize方法的具体用法？C++ Answer::normalize怎么用？C++ Answer::normalize使用的例子？那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类Answer的用法示例。
在下文中一共展示了Answer::normalize方法的1个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C++代码示例。
示例1: main
int main() {
gets(input + 1);
n = strlen(input + 1);
pPower[0] = 1ULL;
for (int i = 1; i <= n; ++i) {
pPower[i] = pPower[i - 1] * P;
prefix[i] = prefix[i - 1] * P + input[i];
}
for (int i = n; i >= 1; --i) {
suffix[i] = suffix[i + 1] * P + input[i];
}
for (int i = n, j = 1; i >= 1; --i) {
int length = n - i + 1;
for (; j <= n && (j < length || hashValue(prefix, j - length, j, length) != hashValue(suffix, n + 1, i, length)); j++);
earliest[i] = j;
}
Answer result;
result.normalize();
for (int i = 1; i <= n; ++i) {
int longest = 1, l = 2, r = min(i, n - i + 1), pos = n + 1;
while (l <= r) {
int mid = l + r >> 1;
if (hashValue(prefix, i - mid, i, mid) == hashValue(suffix, i + mid, i, mid))
longest = mid, l = mid + 1;
else
r = mid - 1;
}
l = i + longest, r = n;
while (l <= r) {
int mid = l + r >> 1;
if (earliest[mid] <= i - longest)
pos = mid, r = mid - 1;
else
l = mid + 1;
}
Answer temp;
temp.v.push_back(Interval(i - longest + 1, i + longest - 1));
if (pos <= n) {
temp.v.push_back(Interval(pos, n));
temp.v.push_back(Interval(earliest[pos] - (n - pos), earliest[pos]));
}
temp.normalize();
update(result, temp);
}
printf("%d\n", (int)result.v.size());
for (int i = 0; i < (int)result.v.size(); ++i)
printf("%d %d\n", result.v[i].first, result.v[i].second - result.v[i].first + 1);
}