Codeforces GoodBye2013 解题报告
2013-12-31 02:41
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第一次能和div1的大神们一起比赛,甚是紧张。只做出来三题,先贴一下吧。
Problem A New Year Candles
题意:有若干新蜡烛,用完两个旧蜡烛就能造一个新蜡烛,每个蜡烛能燃烧1小时。问最多能燃烧多少个小时?
思路:这道题直接爆,虽然有点傻但是比赛的时候保险。
代码如下:
View Code
Problem A New Year Candles
题意:有若干新蜡烛,用完两个旧蜡烛就能造一个新蜡烛,每个蜡烛能燃烧1小时。问最多能燃烧多少个小时?
思路:这道题直接爆,虽然有点傻但是比赛的时候保险。
代码如下:
#include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> #include <utility> #include <queue> #include <stack> #include <vector> #define ll long long #define INF 0x7fffffff #define eps 1E-6 #define LEN 1000100 using namespace std; typedef struct { int x, rat, get; }P; P s[LEN]; int n; bool cmp1(P a, P b){return a.rat<b.rat;} bool cmp2(P a, P b){return a.x<b.x;} int main() { // freopen("in.txt", "r", stdin); while(scanf("%d", &n)!=EOF){ for(int i=0; i<n; i++){ s[i].x = i; scanf("%d", &s[i].rat); } sort(s, s+n, cmp1); int loc = s[0].rat; s[0].get = loc; for(int i=1; i<n; i++){ if(s[i].rat <= loc){ loc++; }else{ loc = s[i].rat; } s[i].get = loc; } sort(s, s+n, cmp2); for(int i=0; i<n; i++){ printf("%d", s[i].get); if(i!=n-1)printf(" "); } printf("\n"); } return 0; }
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