本文整理汇总了C++中vll类的典型用法代码示例。如果您正苦于以下问题:C++ vll类的具体用法?C++ vll怎么用?C++ vll使用的例子?那么恭喜您, 这里精选的类代码示例或许可以为您提供帮助。
在下文中一共展示了vll类的20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,您的评价将有助于我们的系统推荐出更棒的C++代码示例。
示例1: merge
void merge(vll &L, vll &R, vll &A){
ll i=0,j=0,k=0,iL,jL;
iL = L.size();
jL = R.size();
while(i<iL && j<jL ){
if(L[i]<=R[j]){
A[k] = L[i];
i++;
}
else if(R[j]<=L[i]){
A[k] = R[j];
j++;
}
k++;
}
while(i<iL){
A[k] = L[i];
i++;
k++;
}
while(j<jL){
A[k] = R[j];
j++;
k++;
}
return;
}
开发者ID:tushar-rishav,项目名称:Algorithms,代码行数:29,代码来源:merge_sort.cpp
示例2: matmul
vll matmul(const vll &A, const vll &B) {
int m = A.size();
int n = B[0].size();
int x = B.size();
vll C(m, vector<long long>(n, 0));
for (int i=0; i<m; ++i) {
for (int j=0; j<n; ++j) {
for (int k=0; k<x; ++k) {
C[i][j] = (C[i][j] + A[i][k]*B[k][j]) % mod;
}
}
}
return C;
}
开发者ID:alxsoares,项目名称:contest-problem-solutions,代码行数:16,代码来源:TheLongPalindrome.cpp
示例3: main
int main(){
ll n,m,u,v,c;
while(cin >> n >> m && (n||m)){
s=1,t=n+1; N=n+2; pi.assign(N,INF);
cap.assign(N,vll(N,0)); flow.assign(N,vll(N,0)); cost.assign(N,vll(N,0));
vector<ll> U,V,C;
for(int i=0 ; i<m ; i++){
cin >> u >> v >> c;
U.push_back(u); V.push_back(v); C.push_back(c);
}
ll D, K; cin >> D >> K;
for(int i=0 ; i<m ; i++){
u=U[i] , v=V[i], c= C[i];
cost[u][v]=c; cost[v][u]=c;
cap[u][v]=K; cap[v][u]=K;
}
cap[n][t]=D; cost[n+1][t]=0;
mincost=0; maxflow=0; MCMF();
if(maxflow==D) cout << mincost << endl;
else cout << "Impossible." << endl;
}
return 0;
}
开发者ID:ImnIrdst,项目名称:ICPC-Practice-2015-TiZii,代码行数:26,代码来源:UVa+10594+-+Data+Flow+-+AC.cpp
示例4: main
int main(){
ios_base::sync_with_stdio(false);
int T;
ll a, b;
cin >> T;
preprocess();
while(T--){
cin >> a >> b;
if(a == b){
cout << test(a) << '\n';
continue;
}
vll::iterator first = lower_bound(nums.begin(), nums.end(),a), last = lower_bound(nums.begin(), nums.end(),b);
cout << (int)(last - first) + (*last == b) << '\n';
}
}
开发者ID:caioaao,项目名称:cp-solutions,代码行数:17,代码来源:SDSQUARE.cpp
示例5: mul
vll mul(const vll & va, const vll & vb) {
int len = 32 - __builtin_clz(std::max(va.size(), vb.size()) - 1);
len = 1 << (len + 1);
vcd a(len), b(len);
for (int i = 0; i < (int)va.size(); i++) a[i] = cd(va[i], 0);
for (int i = 0; i < (int)vb.size(); i++) b[i] = cd(vb[i], 0);
a = fft(a);
b = fft(b);
for (int i = 0; i < len; i++) {
double real = a[i].real() * b[i].real() - a[i].imag() * b[i].imag();
a[i].imag() = a[i].imag() * b[i].real() + b[i].imag() * a[i].real();
a[i].real() = real;
}
a = fft(a, true);
vll res(len);
for (int i = 0; i < len; i++) res[i] = (long long)(a[i].real() + 0.5);
return res;
}
开发者ID:irajdeep,项目名称:Algorithm-Anthology,代码行数:18,代码来源:4.4.3+FFT+and+Multiplication.cpp
示例6: solve
ll solve(vll dat)
{
int sz=dat.size();
for(int i=1;i<sz;i++)
{
if(i==1)
dat[i]=max(dat[i-1], dat[i]);
else
dat[i]=max(dat[i-1], dat[i]+dat[i-2]);
}
return dat[sz-1];
}
开发者ID:krishnaanaril,项目名称:SPOJ,代码行数:13,代码来源:SAMER08C.cpp
示例7: all
void all( vll& v, vll& l )
{
for( int i = 0; i < v.size(); ++i )
{
vll tmp, comb;
for( vlli j = l.begin(); j != l.end(); ++j )
tmp.push_back( *j + v[ i ] );
for( vlli m = l.begin(), n = tmp.begin(); m != l.end() || n != tmp.end(); )
comb.push_back( n == tmp.end() || m != l.end() && *m <= *n? *m++ : *n++ );
l.swap( comb );
}
}
开发者ID:dibery,项目名称:UVa,代码行数:12,代码来源:12911.cpp
示例8: matexp
vll matexp(vll A, int n) {
int m = A.size();
vll ret = matunit(m);
while (n > 0) {
if (n & 1) {
ret = matmul(ret, A);
}
A = matmul(A, A);
n >>= 1;
}
return ret;
}
开发者ID:alxsoares,项目名称:contest-problem-solutions,代码行数:14,代码来源:TheLongPalindrome.cpp
示例9: matadd
vll matadd(const vll &A, const vll &B) {
int m = A.size();
int n = A[0].size();
vll C(m, vector<long long>(n, 0));
for (int i=0; i<m; ++i) {
for (int j=0; j<n; ++j) {
C[i][j] = A[i][j] + B[i][j];
if (C[i][j] >= mod) {
C[i][j] -= mod;
}
}
}
return C;
}
开发者ID:alxsoares,项目名称:contest-problem-solutions,代码行数:15,代码来源:TheLongPalindrome.cpp
示例10: merge_sort
vll merge_sort(vll S){
vll tempL,tempR;
ll len = S.size();
if(len < 2 )
return S; // array is already sorted. Base case
rep(i,len/2)
tempL.pub(S[i]);
repk(i,len/2,len)
tempR.pub(S[i]);
tempL = merge_sort(tempL);
tempR = merge_sort(tempR);
merge(tempL,tempR,S);
return S;
}
开发者ID:tushar-rishav,项目名称:Algorithms,代码行数:15,代码来源:merge_sort.cpp
示例11: karatsubaMultiply
static vll karatsubaMultiply(const vll &a, const vll &b) {
int n = a.size();
vll res(n + n);
if (n <= 32) {
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
res[i + j] += a[i] * b[j];
return std::move(res);
}
int k = n >> 1;
vll a1(a.begin(), a.begin() + k);
vll a2(a.begin() + k, a.end());
vll b1(b.begin(), b.begin() + k);
vll b2(b.begin() + k, b.end());
vll a1b1 = karatsubaMultiply(a1, b1);
vll a2b2 = karatsubaMultiply(a2, b2);
for (int i = 0; i < k; i++)
a2[i] += a1[i];
for (int i = 0; i < k; i++)
b2[i] += b1[i];
vll r = karatsubaMultiply(a2, b2);
for (int i = 0; i < (int)a1b1.size(); i++)
r[i] -= a1b1[i];
for (int i = 0; i < (int)a2b2.size(); i++)
r[i] -= a2b2[i];
for (int i = 0; i < (int)r.size(); i++)
res[i + k] += r[i];
for (int i = 0; i < (int)a1b1.size(); i++)
res[i] += a1b1[i];
for (int i = 0; i < (int)a2b2.size(); i++)
res[i + n] += a2b2[i];
return std::move(res);
}
开发者ID:condy0919,项目名称:uva-code,代码行数:38,代码来源:254.cpp
示例12: in_lookup
bool in_lookup(const vll &l, long long n) // Lookup table method
{
return std::find(l.begin(), l.end(), n) != l.end();
}
开发者ID:wxv,项目名称:Algorithms,代码行数:4,代码来源:euler_sum_of_powers.cpp
示例13: preprocess
void preprocess(){
nums.push_back(0);
ll sqr = 1, i = 1;
while(sqr <= max_n){
if(test(sqr) && sqr%i == 0) nums.push_back(sqr);
i++; sqr = i*i;
}
}
开发者ID:caioaao,项目名称:cp-solutions,代码行数:8,代码来源:SDSQUARE.cpp
示例14: cal
ll cal() {
vec.push_back(0);
vec.push_back(1);
ll add = 2;
ll a = 2;
for (int i = 2; i < 1000001; ++i) {
vec.push_back(vec[i - 1] + a);
i++;
a += add;
vec.push_back(vec[i - 1] + a);
a += add;
add++;
}
}
开发者ID:ZahraParsaeian,项目名称:Uva_Solved_Problems,代码行数:14,代码来源:11401+Triangle+Counting.cpp
示例15: gen
void gen(void)
{
LL i,j,k;
for(i=1;i<=1005;i++) arr[i]=i*i*i;
for(i=1; i<=1005; i++)
{
for(j=i+1; j<=1005; j++)
{
k=arr[i]+arr[j];
mp[k]++;
if(mp[k]==2) cont.pb(k);
}
}
sort(cont.begin(),cont.end());
return;
}
开发者ID:rofi93,项目名称:UVA,代码行数:16,代码来源:962.cpp
示例16: init
void init()
{
int i=0 , j;
LL pnum=0 , p=1;
sll st;
sll::iterator it;
for(i=0; i<plen; i++)
{
pnum = prm[i];
for(j=2; j<=40; j++)
{
//pnum = power(prm[i] , j);
p=1;
for(int k=1; k<=j; k++)
{
p*=pnum;
if(p > mxhigh) break;
}
if(p < mxhigh)
{
st.insert(p);
}
}
}
//cerr << st.size();
for(it=st.begin() ; it!=st.end() ; ++it)
{
v.push_back(*it);
}
//for(i=0; i<100; i++) cout << v[i] << " ";
//cout << v.size();
//cout << v[v.size()-1];
vlen = v.size();
}
开发者ID:shidhu,项目名称:Competetive-Programming,代码行数:45,代码来源:Dev+Skill+_Mr.+And+Mrs.+A.cpp
示例17: dijkstra
ll dijkstra(int s, int t){
found.assign(N,0); dist.assign(N,INF); f.assign(N,0); dad.assign(N,pii(-1,2));
dist[s]=0; f[s]=INF;
while(s!=-1){
int best=-1; found[s]=true;
for(int k=0 ; k<N; k++){
if(found[k]) continue;
relax(s,k,cap[s][k]-flow[s][k],cost[s][k],1);
relax(s,k,flow[k][s],-1*cost[k][s],-1);
if(best==-1 || dist[k]<dist[best]) best=k;
}
s=best;
}
for(int k=0 ; k<N ; k++){
pi[k]=min(pi[k]+dist[k],INF);
}
return f[t];
}
开发者ID:ImnIrdst,项目名称:ICPC-Practice-2015-TiZii,代码行数:18,代码来源:UVa+10594+-+Data+Flow+-+AC.cpp
示例18: prec
void prec()
{
ll i,j,k,l;
for(i=1; i<INT_MAX; i*=2)
for( j=i; j<INT_MAX; j*=3)
for( k=j; k<INT_MAX; k*=5)
for( l=k; l<INT_MAX; l*=7)
num.pb(l);
sort(all(num));
}
开发者ID:darkprinx,项目名称:ACM-SOLUTION,代码行数:10,代码来源:443-+Humble+Number.cpp
示例19: main
int main ()
{
ll mid;
ll i,j,k,x,ans=0,depend,l,n,ind;
cin>>n>>m>>depend;
fr(i,n)
{
cin>>x;
man.push_back(x);
}
开发者ID:crystal95,项目名称:Competetive-Programming-at-different-platforms,代码行数:12,代码来源:renting.cpp
示例20: count
int count(int n, int k) {
K = k;
T.assign(27, vector<long long>(27, 0));
int start_row = 0;
for (int j=1; j<27; ++j) {
T[start_row][j] = 27-j;
T[++start_row][j] = j;
}
if (n & 1) {
return ((n>1 ? S((n-1)/2) : 0) + S((n+1)/2)) % mod;
} else {
return 2*S(n/2) % mod;
}
}
开发者ID:alxsoares,项目名称:contest-problem-solutions,代码行数:14,代码来源:TheLongPalindrome.cpp
注:本文中的vll类示例由纯净天空整理自Github/MSDocs等源码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。 |
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