题意:给定你有 n 个敌人,你的伤害是 1,给出每个敌人的伤害,和敌人的血量,每一回合你可以攻击一个敌人,并且所有敌人都会攻击你,除非它已经死了,问你最少要多少要消耗多少血量。
析:一个很明显的贪心问题,按照 攻击 / 血量进行排序,然后一个一个的消灭就好了。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #include <list> #include <assert.h> #include <bitset> #include <numeric> #define debug() puts("++++") #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define fi first #define se second #define pb push_back #define sqr(x) ((x)*(x)) #define ms(a,b) memset(a, b, sizeof a) #define sz size() #define be begin() #define ed end() #define pu push_up #define pd push_down #define cl clear() #define lowbit(x) -x&x //#define all 1,n,1 #define FOR(i,n,x) for(int i = (x); i < (n); ++i) #define freopenr freopen("in.in", "r", stdin) #define freopenw freopen("out.out", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e17; const double inf = 1e20; const double PI = acos(-1.0); const double eps = 1e-10; const int maxn = 20 + 5; const int maxm = 700 + 10; const LL mod = 1000000007; const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1}; const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c) { return r >= 0 && r < n && c >= 0 && c < m; } inline int readInt(){ int x; scanf("%d", &x); return x; } struct Node{ int dps, hp; bool operator < (const Node &p) const{ return p.dps * hp < p.hp * dps; } }; Node a[maxn]; int main(){ while(scanf("%d", &n) == 1){ for(int i = 0; i < n; ++i) scanf("%d %d", &a[i].dps, &a[i].hp); sort(a, a + n); LL ans = 0, cnt = 0; for(int i = 0; i < n; ++i){ cnt += a[i].hp; ans += a[i].dps * cnt; } printf("%I64d\n", ans); } return 0; }