树状数组的区间修改和区间查询模板
2017-08-22 12:09
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例题:http://codevs.cn/problem/1082/
#include <iostream>
using namespace std;
#define lowbit(x) ((x) & (-(x)))
long long arrC1[200010], arrC2[200010];
long long iNumTot;
long long GetSum(long long iIndex)
{
long long iAns1 = 0, iAns2 = 0;
for (int i = iIndex; i; i -= lowbit(i))
{
iAns1 += arrC1[i];
iAns2 += arrC2[i];
}
return iIndex * iAns1 - iAns2;
}
void AddNum(int iIndex, long long iDelta)
{
for (int i = iIndex; i <= iNumTot; i += lowbit(i))
{
arrC1[i] += iDelta;
arrC2[i] += (iIndex - 1) * iDelta;
}
}
int main()
{
cin >> iNumTot;
long long iTmp;
for (int i = 1; i <= iNumTot; i++)
{
cin >> iTmp;
AddNum(i, iTmp);
AddNum(i + 1, -iTmp);
}
int iCaseTot; cin >> iCaseTot;
long long iType, a, b, iDelta;
while (iCaseTot--)
{
cin >> iType >> a >> b;
if (a > b)
swap(a, b);
if (iType == 1)
{
cin >> iDelta;
AddNum(a, iDelta);
AddNum(b + 1, -iDelta);
}
else if (iType == 2)
cout << GetSum(b) - GetSum(a - 1) << endl;
}
return 0;
}
#include <iostream>
using namespace std;
#define lowbit(x) ((x) & (-(x)))
long long arrC1[200010], arrC2[200010];
long long iNumTot;
long long GetSum(long long iIndex)
{
long long iAns1 = 0, iAns2 = 0;
for (int i = iIndex; i; i -= lowbit(i))
{
iAns1 += arrC1[i];
iAns2 += arrC2[i];
}
return iIndex * iAns1 - iAns2;
}
void AddNum(int iIndex, long long iDelta)
{
for (int i = iIndex; i <= iNumTot; i += lowbit(i))
{
arrC1[i] += iDelta;
arrC2[i] += (iIndex - 1) * iDelta;
}
}
int main()
{
cin >> iNumTot;
long long iTmp;
for (int i = 1; i <= iNumTot; i++)
{
cin >> iTmp;
AddNum(i, iTmp);
AddNum(i + 1, -iTmp);
}
int iCaseTot; cin >> iCaseTot;
long long iType, a, b, iDelta;
while (iCaseTot--)
{
cin >> iType >> a >> b;
if (a > b)
swap(a, b);
if (iType == 1)
{
cin >> iDelta;
AddNum(a, iDelta);
AddNum(b + 1, -iDelta);
}
else if (iType == 2)
cout << GetSum(b) - GetSum(a - 1) << endl;
}
return 0;
}
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