二分+三分模板
二分:
返回第一个大于val的数
5
1 3 5 6 9
6
>>9
5
1 3 5 7 9
6
>>7
int bsearch(int val){
int l = 0, r = n;
int mid = 0;
while (l <= r){
mid = (l + r)>>1;
if (a[mid] > val){
r = mid - 1;
}
else{
l = mid + 1;
}
}
return a[l];
}
返回第一个大于等于val的值:
int bsearch(int val){
int l = 0, r = n;
int mid = 0;
int ans = 0;
while (l <= r){
mid = (l + r)>>1;
if (a[mid] >= val){
r = mid - 1;
ans = mid;
}
else{
l = mid + 1;
}
}
return a[ans];
}
返回第一个小于val的数
int bsearch(int val){
int l = 0, r = n;
int mid = 0;
while (l <= r){
mid = (l + r)>>1;
if (a[mid] >= val){
r = mid - 1;
}
else{
l = mid + 1;
}
}
return a[r];
}
5
1 3 5 6 9
6
>>5
返回第一个小于等于val的值:
int bsearch(int val){
int l = 0, r = n;
int mid = 0;
int ans = 0;
while (l <= r){
mid = (l + r)>>1;
if (a[mid] > val){
r = mid - 1;
}
else{
l = mid + 1;
ans = mid;
}
}
return a[ans];
}
三分:
double solve(double l,double r)
{
double mid,midmid;
while(dbcmp(r - l) > 0)
{
mid = (l + r)/2.0;
midmid = (r + mid)/2.0;
if (getR(mid) >= getR(midmid)) r = midmid;
else l = mid;
}
return getR(l);
}
{
double mid,midmid;
while(dbcmp(r - l) > 0)
{
mid = (l + r)/2.0;
midmid = (r + mid)/2.0;
if (getR(mid) >= getR(midmid)) r = midmid;
else l = mid;
}
return getR(l);
}


浙公网安备 33010602011771号