9.17百度C++编程题
1.数学题 m*a+n*b>=n*m-k，求a+b最小值，题目说了不等式右边大于零，a和b非负。为了让a+b越小，所以让m，n中的较大值的乘积项越大越好。
#include<bits/stdc++.h>
using namespace std;
int main() {
long long n, m，k;
scanf("%lld %lld %lld", &n, &m, &k);
long long sum = n*m - k;
long long res;
if (m > n) {
res = sum / m + sum % m;
}
else {
res = sum / n + sum % n;
}
printf("%lld", res);
}
2.安排任务，给出各个任务的所需时间和deadline，看是否有任务会超时，模拟一下
#include<bits/stdc++.h>
using namespace std;
bool cmp(pair<int, int> a, pair<int, int>b) {
if (a.second != b.second)return a.second < b.second;
return a.first < b.first;
}
int main() {
int t,n, a, b;
scanf("%d", &t);
while (t--) {
scanf("%d", &n);
vector<pair<int, int>> v;
for (int i = 0; i < n; i++) {
scanf("%d%d", &a, &b);
v.push_back({ a,b });
}
sort(v.begin(), v.end(), cmp);
int cur = 0;
bool flag = true;
for (auto it : v) {
cur += it.first;
if (cur > it.second) {
flag = false;
break;
}
}
printf(flag ? "Yes\n" : "No\n");
}
return 0;
}
3.二人博弈。用dfs遍历了两次，找到两人能访问的中间结点，和各自一边的节点。中间结点数为偶数，两人平分；为奇数，牛牛多取一个。牛牛先手，胜利条件是牛牛能取的点要比牛妹多 #include<bits/stdc++.h>
using namespace std;
int t, n, a, b;
vector<int> v[100005];
bool vis1[100005], vis2[100005];
void dfs1(int index) {
if (index == n)return;
vis1[index] = true;
for (int i = 0; i < v[index].size(); i++) {
if(vis1[v[index][i]]==false)dfs1(v[index][i]);
}
}
void dfs2(int index) {
if (index == 1)return;
vis2[index] = true;
for (int i = 0; i < v[index].size(); i++) {
if (vis2[v[index][i]] == false) dfs2(v[index][i]);
}
}
int main() {
scanf("%d", &t);
while (t--) {
scanf("%d", &n);
for (int i = 0; i < 100005; i++) {
v[i].clear();
}
fill(vis1, vis1 + 100005, 0);
fill(vis2, vis2 + 100005, 0);
for (int i = 0; i < n-1; i++) {
scanf("%d%d", &a, &b);
v[a].push_back(b);
v[b].push_back(a);
}
int res1 = 0, res2 = 0;
dfs1(1);
dfs2(n);
for (int i = 2; i < n; i++) {
if (vis1[i]) {
if (vis2[i]) {
res1++;
res2++;
}else {
res1 += 2;
}
}else {
if (vis1[2]) {
res2 += 2;
}
}
}
res1 = res1 / 2 + res1 % 2) ;
res2 = res2 / 2;
if (res1 > res2)printf("niuniu\n");
else printf("niumei\n");
}
return 0;
} 
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