标签：mod reverse rev sum 快速 算法 加法 quicksort return
排序：
1.冒泡排序(什么？你也是冒泡人？):
void bubble_Sort(int* arr, int n) {
for (int i = 0; i<n-1; ++i)
for (int j = 0; j<n-i-1; ++j)
if (arr[j] > arr[j+1])
swap(arr[j], arr[j+1]);
}
2.快速排序:
void quickSort(int* arr, int low, int high) {
if (low >= high)
return;
int kase = low;
for (int i = low; i<high; i++)
if (arr[i] < arr[high])
swap(arr[kase++], arr[i]);
swap(arr[kase], arr[high]);
quickSort(arr, low, kase-1);
quickSort(arr, kase+1, high);
}
3.归并排序：
void mergeSort(int arr[], int temp[], int low, int high) {
if (low >= high)
return;
int len = high-low, mid = len/2 + low;
int start1 = low, end1 = mid, start2 = mid+1, end2 = high;
mergeSort(arr, temp, start1, end1);
mergeSort(arr, temp, start2, end2);
int index = low;
while(start1 <= end1 && start2 <= end2)
temp[index++] = arr[start1] <= arr[start2] ? arr[start1++] : arr[start2++];
while(start1 <= end1)
temp[index++] = arr[start1++];
while(start2 <= end2)
temp[index++] = arr[start2++];
for (int i = low; i<=high; i++)
arr[i] = temp[i];
}
4.待续....
快速幂&&快速乘：
1.快速幂：
ll quickPow(ll x, ll y) {
if (!y)
return 1;
ll res = quickPow(x, y >> 1);
res = res*res%MOD;
if (y&1)
res = res*x%MOD;
return res;
}
2.快速乘：
ll solve(ll n, ll m) {
if (!n || !m)
return 0;
ll res = solve(n, m>>1);
res = res<<1 % MOD;
if (m&1)
res = (res+n) % MOD;
return res;
}
高精度：
1.加法
void sum() {
int len1 = strlen(num1);
int len2 = strlen(num2);
reverse(num1, num1+len1);
reverse(num2, num2+len2);
int kase = 0;
while(kase < len1 && kase < len2) {
res[kase+1] = (res[kase] + num1[kase] + num2[kase] - ‘0‘*2)/10;
res[kase] = (res[kase] + (num1[kase] + num2[kase] - ‘0‘*2))%10 + ‘0‘;
++kase;
}
while(kase < len1) {
res[kase+1] = (res[kase] + num1[kase] - ‘0‘)/10;
res[kase] = (res[kase] + num1[kase] - ‘0‘)%10 + ‘0‘;
++kase;
}
while(kase < len2) {
res[kase+1] = (res[kase] + num2[kase] - ‘0‘)/10;
res[kase] = (res[kase] + num2[kase] - ‘0‘)%10 + ‘0‘;
++kase;
}
res[kase] += ‘0‘;
for (int i = kase; i>=0; i--) {
if (res[kase] == ‘0‘)
continue;
cout << res[i];
}
cout << endl;
fill(num1, num1+maxn, 0);
fill(num2, num2+maxn, 0);
fill(res, res+maxn, 0);
}
尺取法：
标签：mod reverse rev sum 快速 算法 加法 quicksort return
原文地址：https://www.cnblogs.com/shuitiangong/p/12063528.html
