P=[30.1
13369329273
1,23.7941905551386,25.9274232393425,3.99093290154067,1.07094210534649,
15.60407785
73771,-6.74973974521525,17.9316192151211,17.7751441978811,17.0367626807968,6.44145507097795,18.6368049401464,18.0550487545728,17.8305848999805,-9.04597306585009,4.25781023813848,14.0411117370762,25.6537565169466,32.6530114743632,11.4232297257833,22.1767312287518,5.35814454198668,5.54986503126230,19.8010231090071,20.8737867077591,22.6667971860760,24.0571632300461,-1.17853515692208,21.3911350310525,11.6330942835818,7.09484524327073,10.8448856959349,30.9127254672511,19.0396418491542,37.7801710882361,39.8531941093693,34.2038053560476,28.6714863290273,17.36
13878840909
,42.9072554206643,32.1271103281040,16.1438792137387,12.9988226694107,
15.13890338
10301,12.9083762904698,25.4258971642663,-7.38764858445742,18.5848932630656,31.1616370060378,27.
18080077766
69,39.8759677718025,14.7808600539316,13.1429976155788,16.4862287949336,11.2829313588549,10.5269580000565,-2.24368440843136,
15.67537201
45639,-7.95459199158182,22.8919745314062,8.20120623586929,
15.25877285
25311,17.2934463414763,10.2555682439300,-8.92908481276116,13.1439197791044,19.2399648766200,
15.89495899
46201,-12.7709471106909,-8.5
15945567022
17,0.669
18164393362
6,27.4478553286513,32.7732639885067,19.2970930617401,19.8499399698847,-1.42930324043134,0.882469009504931,11.78
13104904510
,13.4146569650138,12.1299990647653,6.19010367749992,13.2217864231353,23.1707180459207,8.76878162712805,18.6035876418327,22.2293396599091,21.4504949833715,31.4599792538974,37.2878064624533,1.82260861337259,-2.94178514920019,-11.7085852367391,33.5960558316278,-18.2166510154666,17.3
15226144137
3,17.0298307564903,13.7470827204786,3.12
13394535981
8,-13.9378864418342,12.0
13666368454
0,];
P1=-P;
b=0;
sum1=0;
sum2=0;
x=0;
P3=sort(P1);
for i=1:100
if P3(i)<=0 && P3(i+1)>=0
x=i;
belta=(100-i)/100;
end
end
for i=1:100
if P1(i)>=0
sum1=sum1+P1(i);
else sum2=sum2+P1(i);
end
end

P只是赋值100个数 啊 没有很复杂

既然极为简单，你为何还不会做？