Codeforces Beta Round #4 (Div. 2 Only) B. Before an Exam
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继续刷水题,不能再错了!!
题目大意:
要求在n天内学习t小时,给出每一天可以学习的最小时间和最大时间。问能不能实现,若可以输出时间表。
解题思路:
水题一个,保证在区间范围内的情况下直接贪心。
下面是代码:
#include <set>#include <map>#include <queue>#include <math.h>#include <vector>#include <string>#include <stdio.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <cctype>#include <algorithm>#define eps 1e-10#define pi acos(-1.0)#define inf 107374182#define inf64 1152921504606846976#define lc l,m,tr<<1#define rc m + 1,r,tr<<1|1#define zero(a) fabs(a)<eps#define iabs(x) ((x) > 0 ? (x) : -(x))#define clear1(A, X, SIZE) memset(A, X, sizeof(A[0]) * (min(SIZE,sizeof(A))))#define clearall(A, X) memset(A, X, sizeof(A))#define memcopy1(A , X, SIZE) memcpy(A , X ,sizeof(X[0])*(SIZE))#define memcopyall(A, X) memcpy(A , X ,sizeof(X))#define max( x, y ) ( ((x) > (y)) ? (x) : (y) )#define min( x, y ) ( ((x) < (y)) ? (x) : (y) )using namespace std;struct node{ int min1,max1;}da[35];int main(){ int d,t,uplim=0,downlim=0; scanf("%d%d",&d,&t); for(int i=0;i<d;i++) { scanf("%d%d",&da[i].min1,&da[i].max1); downlim+=da[i].min1; uplim+=da[i].max1; } if(uplim<t||downlim>t) { puts("NO"); } else { puts("YES"); t-=downlim; for(int i=0;i<d;i++) { if(t>=da[i].max1-da[i].min1) { printf("%d ",da[i].max1); t-=da[i].max1-da[i].min1; } else if(t==0) { printf("%d ",da[i].min1); } else if(t<da[i].max1-da[i].min1) { printf("%d ",t+da[i].min1); t=0; } } } return 0;}
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