本文整理汇总了C++中vi::front方法的典型用法代码示例。如果您正苦于以下问题：C++ vi::front方法的具体用法？C++ vi::front怎么用？C++ vi::front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vi的用法示例。
在下文中一共展示了vi::front方法的5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: binary_search
int binary_search(int start, int end) {
while(start <= end){
if(abs(start - end) == 1) return vec[start] + getR();
int mid = (start + end)/2;
if(vec[mid] == getN(vec.front(), mid, getR())){
start = mid;
} else {
end = mid;
}
}
}示例2: solve
void solve() {
arr.clear();
p.clear();
/*
cin >> K >> N;
arr = vi(K);
forn(i, K)
cin >> arr[i];
*/
cin >> N >> K;
int64 b, c, r;
arr = vi(K);
cin >> arr[0] >> b >> c >> r;
forab(i, 1, K-1)
arr[i] = (arr[i-1]*b + c) % r;
forn(i, K)
write(arr[i]);
cout << endl;
sort(all(arr));
uni.pb(arr.front());
p.pb(mp(arr[0], arr[0]));
forn(i, K) {
if(arr[i] != uni.back())
uni.pb(arr[i]);
if(arr[i] == p.back().second)
;
else if(arr[i] == p.back().second+1)
p.back().second++;
else
p.pb(mp(arr[i], arr[i]));
}
forn(i, K)
write(arr[i]);
cout << endl;
forn(i, uni.size())
write(uni[i]);
cout << endl;
tr(it, p)
Pf("(%d, %d) ", it->first, it->second);
cout << endl;
int M = N;
forab(i, K, N) {
N = i;
Pf("arr[%d] = %d\n", N, binarySearch());
}示例3: main
int main (){
scanf("%d%d", &n, &k);
for (int i = 0; i < n * k / 2; i++){
scanf("%d%d", &x, &y);
a[--x][--y] = a[y][x] = 1;
st[x]++;st[y]++;
if (x > y)swap(x,y);
edge.push_back(make_pair(x,y));
}
for (int i = 0; i < n; i++)
if (st[i]){
euler(i);
for (int j = 1; j < eu.size(); j++)
g[eu[j - 1]].push_back(eu[j]);
if (eu.back() != eu.front())
g[eu.back()].push_back(eu.front());
eu.clear();
}
mt.assign(n, -1);
for (int i = 0; i < n; i++){
u.assign(n, 0);
dfs(i);
}
for (int i = 0; i < n; i++)
ans += mt[i] != -1;
if (ans == n){
puts("YES");
for (int i = 0; i < n; i++){
x = i, y = mt[i];
if (x > y)swap(x,y);
for (int j = 0; j < edge.size(); j++)
if (x == edge[j].first && y == edge[j].second){
printf("%d\n", j + 1);break;}
}
} else puts("NO");
return 0;
}示例4: getR
int getR(){
return (vec.back() - vec.front())/(vec.size());
}示例5: main
int main() {
#ifdef DEBUG
time_t startt = clock();
#endif
isprime.set();
isprime[0] = isprime[1] = 0;
for(int i = 0; i<20015; i++)
if(isprime[i]) {
for(int j = i*i; j<20015; j+= i)
isprime[j] = 0;
}
scanf("%d", &n);
p.assign(n+2, 0);
q.assign(n+2, 0);
int cnt[] = {0, 0};
s = n; t = n+1;
memset(res, 0, sizeof res);
for(int i = 0; i<n; i++) {
scanf("%d", a+i);
if(a[i] & 1) {
cnt[0]++;
res[s][i] = 2;
}
else {
res[i][t] = 2;
cnt[1]++;
}
}
if(cnt[0] != cnt[1]) {
puts("Impossible");
return 0;
}
for(int i = 1; i<n; i++)
for(int j = 0; j<i; j++)
if(isprime[a[i]+a[j]]) {
if(a[i] & 1)
res[i][j] = 1;
else
res[j][i] = 1;
}
int flow = 0;
n += 2;
while(1) {
f = 0;
vis.reset();
queue< int > q;
p.assign(n, -1);
q.push(s);
vis[s] = 1;
while(!q.empty()) {
int u = q.front(); q.pop();
if(u == t) break;
for(int i = 0; i<n; i++)
if(!vis[i] && res[u][i] > 0)
p[i] = u, vis[i] = 1, q.push(i);
}
augment(t, 1000000);
if(f == 0) break;
flow += f;
}
if(flow != (cnt[0]<<1)) {
puts("Impossible");
return 0;
}
vis.set();
n -= 2;
for(int i = 0; i<n; i++) if(vis[i] && (a[i] & 1)) {
int v = i;
ans.push_back(vi());
while(vis[v]) {
ans.back().push_back(v);
vis[v] = 0;
for(int j = 0; j<n; j++) {
if(vis[j] && ((a[j] & 1) == 0) && res[j][v]) {
v = j;
break;
}
}
ans.back().push_back(v);
vis[v] = 0;
for(int j = 0; j<n; j++)
if(vis[j] && (a[j] & 1) && res[v][j]) {
v = j;
break;
}
}
}
printf("%d\n", ans.size());
for(int i = 0; i<ans.size(); i++) {
printf("%d", ans[i].size());
for(int j = 0; j< ans[i].size(); j++)
printf(" %d", ans[i][j] + 1);
printf("\n");
}
#ifdef DEBUG
cerr << "Running time : " << (double)(clock()-startt) * 1000 / CLOCKS_PER_SEC << " ms" << endl;
int T = time(NULL)-1491365861;
cerr << "Time : " << T/60 << " minutes " << T%60 << " secs" << endl;
#endif
return 0;
//.........这里部分代码省略.........